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Chapter 3: Electricity

NCERT Solutions for CBSE Class 10 Physics — 817 solved questions with detailed explanations.

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Important Formulas

Statement

Ohm's Law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

V = IR

Where V = potential difference (volts), I = current (amperes), R = resistance (ohms, Ω).

Understanding Resistance

Resistance is the property of a conductor that opposes the flow of current. Think of it as friction for electricity. A thicker wire has less resistance (more room for electrons), while a thinner wire has more resistance.

The SI unit of resistance is the ohm (Ω). 1 ohm is the resistance of a conductor through which 1 ampere of current flows when 1 volt of potential difference is applied across it.

Ohm's Law Graph

When we plot a graph of V (on x-axis) against I (on y-axis) for an ohmic conductor, we get a straight line passing through the origin. The slope of this line gives 1/R (or the reciprocal of resistance).

Note: Not all conductors obey Ohm's law. Components like diodes and transistors are non-ohmic — their V-I graphs are not straight lines.

Factors Affecting Resistance

The resistance of a conductor depends on:

Combining these factors:

R = ρl/A

Resistivity (ρ)

The constant ρ (rho) is called resistivity or specific resistance. It depends only on the material and temperature, not on the dimensions of the conductor.

SI unit of resistivity: Ω·m (ohm-metre)

Resistivity Values

Series Combination

When resistors are connected end-to-end so that the same current flows through each, they are said to be in series.

Key Properties of Series Circuits

R_eq = R₁ + R₂ + R₃ + ...

Derivation

Since V = V₁ + V₂ + V₃ and V = IR:

IR_eq = IR₁ + IR₂ + IR₃

Dividing both sides by I:

R_eq = R₁ + R₂ + R₃

Important Note

In a series circuit, the equivalent resistance is always greater than the largest individual resistance. If one component breaks, the entire circuit breaks (like old-style Christmas lights).

Parallel Combination

When resistors are connected between the same two points so that the same potential difference is applied across each, they are in parallel.

Key Properties of Parallel Circuits

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + ...

Derivation

Since I = I₁ + I₂ + I₃ and I = V/R:

V/R_eq = V/R₁ + V/R₂ + V/R₃

Dividing both sides by V:

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃

For Two Resistors in Parallel

R_eq = (R₁ × R₂) / (R₁ + R₂)

Important Note

In a parallel circuit, the equivalent resistance is always less than the smallest individual resistance. If one component breaks, the others continue working (this is how household wiring works).

Joule's Law of Heating

When current flows through a resistor, electrical energy is converted to heat energy. This is called the heating effect of electric current (or Joule heating).

The heat produced (H) is given by:

H = I²Rt

Where H = heat in joules, I = current in amperes, R = resistance in ohms, t = time in seconds.

Derivation

Work done by the source: W = VIt

Using Ohm's law (V = IR): W = (IR)(I)(t) = I²Rt

This work is converted entirely into heat: H = I²Rt

Other Forms

Applications

Electric Power

Power is the rate at which electrical energy is consumed or dissipated.

P = VI = I²R = V²/R

SI unit: watt (W). 1 watt = 1 volt × 1 ampere = 1 joule per second.

1 kilowatt (kW) = 1000 W

Electric Energy

Energy consumed = Power × Time

E = P × t = VIt

SI unit: joule (J)

Commercial Unit of Energy

The commercial unit of electrical energy is the kilowatt-hour (kWh), commonly called a "unit" on electricity bills.

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J = 3.6 MJ

Example: If a 100 W bulb runs for 10 hours, energy consumed = 100 × 10 = 1000 Wh = 1 kWh = 1 "unit".

Solved Questions

Q1. The SI unit of electric current is:

Difficulty: Easy · Topic: Electric Current

Solution

The SI unit of electric current is the ampere (A), named after André-Marie Ampère. 1 ampere is defined as the flow of 1 coulomb of charge per second. Volt is the unit of potential difference, ohm is the unit of resistance, and coulomb is the unit of charge.

Q2. A conductor has a potential difference of 28 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{28}{0.5} = 56.0\;\Omega\)

Q3. A conductor has a potential difference of 46 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{46}{0.25} = 184.0\;\Omega\)

Q4. A conductor has a potential difference of 39 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{39}{0.25} = 156.0\;\Omega\)

Q5. A conductor has a potential difference of 37 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{37}{5.0} = 7.4\;\Omega\)

Q6. A conductor has a potential difference of 29 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{29}{2.0} = 14.5\;\Omega\)

Q7. A conductor has a potential difference of 3 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{3}{0.5} = 6.0\;\Omega\)

Q8. A conductor has a potential difference of 21 V across it and a current of 2.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{21}{2.5} = 8.4\;\Omega\)

Q9. A conductor has a potential difference of 8 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{8}{3.0} = 2.67\;\Omega\)

Q10. A conductor has a potential difference of 9 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{9}{2.0} = 4.5\;\Omega\)

Q11. A conductor has a potential difference of 49 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{49}{3.0} = 16.33\;\Omega\)

Q12. A conductor has a potential difference of 18 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{18}{0.25} = 72.0\;\Omega\)

Q13. A conductor has a potential difference of 20 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{20}{0.25} = 80.0\;\Omega\)

Q14. A conductor has a potential difference of 46 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{46}{3.0} = 15.33\;\Omega\)

Q15. A conductor has a potential difference of 14 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{14}{2.0} = 7.0\;\Omega\)

Q16. A conductor has a potential difference of 5 V across it and a current of 1.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{5}{1.0} = 5.0\;\Omega\)

Q17. A conductor has a potential difference of 35 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{35}{3.0} = 11.67\;\Omega\)

Q18. A conductor has a potential difference of 48 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{48}{5.0} = 9.6\;\Omega\)

Q19. A conductor has a potential difference of 41 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{41}{0.25} = 164.0\;\Omega\)

Q20. A conductor has a potential difference of 28 V across it and a current of 1.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{28}{1.0} = 28.0\;\Omega\)

Q21. A conductor has a potential difference of 12 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{12}{5.0} = 2.4\;\Omega\)

Q22. A conductor has a potential difference of 8 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{8}{0.5} = 16.0\;\Omega\)

Q23. A conductor has a potential difference of 28 V across it and a current of 1.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{28}{1.5} = 18.67\;\Omega\)

Q24. A conductor has a potential difference of 37 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{37}{3.0} = 12.33\;\Omega\)

Q25. A conductor has a potential difference of 8 V across it and a current of 2.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{8}{2.5} = 3.2\;\Omega\)

Q26. A conductor has a potential difference of 21 V across it and a current of 1.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{21}{1.0} = 21.0\;\Omega\)

Q27. A conductor has a potential difference of 37 V across it and a current of 0.1 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{37}{0.1} = 370.0\;\Omega\)

Q28. A conductor has a potential difference of 30 V across it and a current of 4.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{30}{4.0} = 7.5\;\Omega\)

Q29. A conductor has a potential difference of 22 V across it and a current of 1.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{22}{1.0} = 22.0\;\Omega\)

Q30. A conductor has a potential difference of 24 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{24}{0.25} = 96.0\;\Omega\)

Q31. A conductor has a potential difference of 48 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{48}{2.0} = 24.0\;\Omega\)

Q32. Two resistors \(R_1 = 11\;\Omega\) and \(R_2 = 39\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 11 + 39 = 50\;\Omega\)

Q33. Two resistors \(R_1 = 4\;\Omega\) and \(R_2 = 42\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 4 + 42 = 46\;\Omega\)

Q34. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 3\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 25 + 3 = 28\;\Omega\)

Q35. Two resistors \(R_1 = 42\;\Omega\) and \(R_2 = 46\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 42 + 46 = 88\;\Omega\)

Q36. Two resistors \(R_1 = 30\;\Omega\) and \(R_2 = 18\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 30 + 18 = 48\;\Omega\)

Q37. Two resistors \(R_1 = 3\;\Omega\) and \(R_2 = 31\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 3 + 31 = 34\;\Omega\)

Q38. Two resistors \(R_1 = 49\;\Omega\) and \(R_2 = 4\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 49 + 4 = 53\;\Omega\)

Q39. Two resistors \(R_1 = 5\;\Omega\) and \(R_2 = 37\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 5 + 37 = 42\;\Omega\)

Q40. Two resistors \(R_1 = 17\;\Omega\) and \(R_2 = 24\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 17 + 24 = 41\;\Omega\)

Q41. Two resistors \(R_1 = 47\;\Omega\) and \(R_2 = 14\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 47 + 14 = 61\;\Omega\)

Q42. Two resistors \(R_1 = 7\;\Omega\) and \(R_2 = 37\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 7 + 37 = 44\;\Omega\)

Q43. Two resistors \(R_1 = 48\;\Omega\) and \(R_2 = 43\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 48 + 43 = 91\;\Omega\)

Q44. Two resistors \(R_1 = 24\;\Omega\) and \(R_2 = 28\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 24 + 28 = 52\;\Omega\)

Q45. Two resistors \(R_1 = 19\;\Omega\) and \(R_2 = 16\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 19 + 16 = 35\;\Omega\)

Q46. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 35\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 25 + 35 = 60\;\Omega\)

Q47. Two resistors \(R_1 = 44\;\Omega\) and \(R_2 = 38\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 44 + 38 = 82\;\Omega\)

Q48. Two resistors \(R_1 = 4\;\Omega\) and \(R_2 = 26\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 4 + 26 = 30\;\Omega\)

Q49. Two resistors \(R_1 = 24\;\Omega\) and \(R_2 = 19\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 24 + 19 = 43\;\Omega\)

Q50. Two resistors \(R_1 = 2\;\Omega\) and \(R_2 = 50\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 2 + 50 = 52\;\Omega\)

Q51. Two resistors \(R_1 = 49\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 49 + 13 = 62\;\Omega\)

Q52. Two resistors \(R_1 = 11\;\Omega\) and \(R_2 = 47\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 11 + 47 = 58\;\Omega\)

Q53. Two resistors \(R_1 = 19\;\Omega\) and \(R_2 = 6\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 19 + 6 = 25\;\Omega\)

Q54. Two resistors \(R_1 = 17\;\Omega\) and \(R_2 = 39\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 17 + 39 = 56\;\Omega\)

Q55. Two resistors \(R_1 = 12\;\Omega\) and \(R_2 = 4\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 12 + 4 = 16\;\Omega\)

Q56. Two resistors \(R_1 = 39\;\Omega\) and \(R_2 = 7\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 39 + 7 = 46\;\Omega\)

Q57. Two resistors \(R_1 = 34\;\Omega\) and \(R_2 = 15\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 34 + 15 = 49\;\Omega\)

Q58. Two resistors \(R_1 = 7\;\Omega\) and \(R_2 = 2\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 7 + 2 = 9\;\Omega\)

Q59. Two resistors \(R_1 = 21\;\Omega\) and \(R_2 = 2\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 21 + 2 = 23\;\Omega\)

Q60. Two resistors \(R_1 = 41\;\Omega\) and \(R_2 = 7\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 41 + 7 = 48\;\Omega\)

Q61. Two resistors \(R_1 = 21\;\Omega\) and \(R_2 = 47\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 21 + 47 = 68\;\Omega\)

Q62. A current of 5.0 A flows through a resistance of 45 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 45 = 225.0\) V

Q63. A current of 1.5 A flows through a resistance of 25 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 25 = 37.5\) V

Q64. A current of 0.25 A flows through a resistance of 25 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 25 = 6.25\) V

Q65. A current of 0.2 A flows through a resistance of 30 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 30 = 6.0\) V

Q66. A current of 1.0 A flows through a resistance of 9 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 9 = 9.0\) V

Q67. A current of 3.0 A flows through a resistance of 2 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 3.0 \times 2 = 6.0\) V

Q68. A current of 3.0 A flows through a resistance of 37 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 3.0 \times 37 = 111.0\) V

Q69. A current of 1.5 A flows through a resistance of 10 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 10 = 15.0\) V

Q70. A current of 1.5 A flows through a resistance of 61 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 61 = 91.5\) V

Q71. A current of 5.0 A flows through a resistance of 77 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 77 = 385.0\) V

Q72. A current of 5.0 A flows through a resistance of 37 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 37 = 185.0\) V

Q73. A current of 2.5 A flows through a resistance of 2 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.5 \times 2 = 5.0\) V

Q74. A current of 5.0 A flows through a resistance of 97 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 97 = 485.0\) V

Q75. A current of 0.25 A flows through a resistance of 50 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 50 = 12.5\) V

Q76. A current of 1.5 A flows through a resistance of 100 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 100 = 150.0\) V

Q77. A current of 4.0 A flows through a resistance of 27 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 4.0 \times 27 = 108.0\) V

Q78. A current of 0.5 A flows through a resistance of 7 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.5 \times 7 = 3.5\) V

Q79. A current of 0.5 A flows through a resistance of 70 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.5 \times 70 = 35.0\) V

Q80. A current of 3.0 A flows through a resistance of 28 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 3.0 \times 28 = 84.0\) V

Q81. A current of 4.0 A flows through a resistance of 26 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 4.0 \times 26 = 104.0\) V

Q82. A current of 0.25 A flows through a resistance of 91 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 91 = 22.75\) V

Q83. A current of 3.0 A flows through a resistance of 65 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 3.0 \times 65 = 195.0\) V

Q84. A current of 0.2 A flows through a resistance of 5 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 5 = 1.0\) V

Q85. A current of 2.0 A flows through a resistance of 35 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.0 \times 35 = 70.0\) V

Q86. A current of 0.25 A flows through a resistance of 41 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 41 = 10.25\) V

Q87. A current of 1.0 A flows through a resistance of 48 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 48 = 48.0\) V

Q88. A current of 0.5 A flows through a resistance of 77 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.5 \times 77 = 38.5\) V

Q89. A current of 0.25 A flows through a resistance of 61 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 61 = 15.25\) V

Q90. A current of 2.0 A flows through a resistance of 62 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.0 \times 62 = 124.0\) V

Q91. A current of 0.2 A flows through a resistance of 28 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 28 = 5.6\) V

Q92. A 3000 W device runs for 10 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 10\text{ h} = 30.0\) kWh

Q93. A 1500 W device runs for 3 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 3\text{ h} = 4.5\) kWh

Q94. A 200 W device runs for 2 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 200\text{ W} \times 2\text{ h} = 0.4\) kWh

Q95. A 3000 W device runs for 1 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 1\text{ h} = 3.0\) kWh

Q96. A 1500 W device runs for 12 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 12\text{ h} = 18.0\) kWh

Q97. A 750 W device runs for 7 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 750\text{ W} \times 7\text{ h} = 5.25\) kWh

Q98. A 200 W device runs for 3 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 200\text{ W} \times 3\text{ h} = 0.6\) kWh

Q99. A 2000 W device runs for 7 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 7\text{ h} = 14.0\) kWh

Q100. A 200 W device runs for 1 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 200\text{ W} \times 1\text{ h} = 0.2\) kWh

Q101. A 1500 W device runs for 11 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 11\text{ h} = 16.5\) kWh

Q102. A 750 W device runs for 3 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 750\text{ W} \times 3\text{ h} = 2.25\) kWh

Q103. A 500 W device runs for 1 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 500\text{ W} \times 1\text{ h} = 0.5\) kWh

Q104. A 2000 W device runs for 4 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 4\text{ h} = 8.0\) kWh

Q105. A 1500 W device runs for 7 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 7\text{ h} = 10.5\) kWh

Q106. A 1000 W device runs for 4 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 4\text{ h} = 4.0\) kWh

Q107. A 1500 W device runs for 5 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 5\text{ h} = 7.5\) kWh

Q108. A 100 W device runs for 7 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 100\text{ W} \times 7\text{ h} = 0.7\) kWh

Q109. A 500 W device runs for 8 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 500\text{ W} \times 8\text{ h} = 4.0\) kWh

Q110. A 500 W device runs for 6 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 500\text{ W} \times 6\text{ h} = 3.0\) kWh

Q111. A 750 W device runs for 11 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 750\text{ W} \times 11\text{ h} = 8.25\) kWh

Q112. A 3000 W device runs for 6 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 6\text{ h} = 18.0\) kWh

Q113. A 1000 W device runs for 11 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 11\text{ h} = 11.0\) kWh

Q114. A 2000 W device runs for 8 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 8\text{ h} = 16.0\) kWh

Q115. A 750 W device runs for 12 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 750\text{ W} \times 12\text{ h} = 9.0\) kWh

Q116. A 2000 W device runs for 5 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 5\text{ h} = 10.0\) kWh

Q117. A 2000 W device runs for 12 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 12\text{ h} = 24.0\) kWh

Q118. A 1000 W device runs for 9 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 9\text{ h} = 9.0\) kWh

Q119. A conductor has a potential difference of 44 V across it and a current of 4.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{44}{4.0} = 11.0\;\Omega\)

Q120. A conductor has a potential difference of 42 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{42}{5.0} = 8.4\;\Omega\)

Q121. A conductor has a potential difference of 27 V across it and a current of 1.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{27}{1.0} = 27.0\;\Omega\)

Q122. A conductor has a potential difference of 16 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{16}{0.25} = 64.0\;\Omega\)

Q123. A conductor has a potential difference of 34 V across it and a current of 2.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{34}{2.5} = 13.6\;\Omega\)

Q124. A conductor has a potential difference of 20 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{20}{2.0} = 10.0\;\Omega\)

Q125. A conductor has a potential difference of 6 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{6}{0.25} = 24.0\;\Omega\)

Q126. A conductor has a potential difference of 21 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{21}{0.5} = 42.0\;\Omega\)

Q127. A conductor has a potential difference of 7 V across it and a current of 1.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{7}{1.5} = 4.67\;\Omega\)

Q128. A conductor has a potential difference of 16 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{16}{3.0} = 5.33\;\Omega\)

Q129. A conductor has a potential difference of 42 V across it and a current of 1.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{42}{1.5} = 28.0\;\Omega\)

Q130. A conductor has a potential difference of 23 V across it and a current of 0.1 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{23}{0.1} = 230.0\;\Omega\)

Q131. A conductor has a potential difference of 22 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{22}{0.25} = 88.0\;\Omega\)

Q132. A conductor has a potential difference of 25 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{25}{0.5} = 50.0\;\Omega\)

Q133. A conductor has a potential difference of 21 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{21}{5.0} = 4.2\;\Omega\)

Q134. A conductor has a potential difference of 26 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{26}{0.25} = 104.0\;\Omega\)

Q135. A conductor has a potential difference of 23 V across it and a current of 1.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{23}{1.5} = 15.33\;\Omega\)

Q136. A conductor has a potential difference of 43 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{43}{2.0} = 21.5\;\Omega\)

Q137. A conductor has a potential difference of 19 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{19}{5.0} = 3.8\;\Omega\)

Q138. A conductor has a potential difference of 3 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{3}{5.0} = 0.6\;\Omega\)

Q139. A conductor has a potential difference of 33 V across it and a current of 4.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{33}{4.0} = 8.25\;\Omega\)

Q140. A conductor has a potential difference of 32 V across it and a current of 1.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{32}{1.5} = 21.33\;\Omega\)

Q141. A conductor has a potential difference of 27 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{27}{2.0} = 13.5\;\Omega\)

Q142. A conductor has a potential difference of 7 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{7}{0.25} = 28.0\;\Omega\)

Q143. A conductor has a potential difference of 19 V across it and a current of 2.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{19}{2.5} = 7.6\;\Omega\)

Q144. A conductor has a potential difference of 39 V across it and a current of 0.1 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{39}{0.1} = 390.0\;\Omega\)

Q145. A conductor has a potential difference of 16 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{16}{2.0} = 8.0\;\Omega\)

Q146. A conductor has a potential difference of 40 V across it and a current of 0.1 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{40}{0.1} = 400.0\;\Omega\)

Q147. A conductor has a potential difference of 2 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{2}{5.0} = 0.4\;\Omega\)

Q148. A conductor has a potential difference of 13 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{13}{0.5} = 26.0\;\Omega\)

Q149. A conductor has a potential difference of 44 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{44}{0.5} = 88.0\;\Omega\)

Q150. A conductor has a potential difference of 14 V across it and a current of 0.1 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{14}{0.1} = 140.0\;\Omega\)

Q151. A conductor has a potential difference of 30 V across it and a current of 0.1 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{30}{0.1} = 300.0\;\Omega\)

Q152. A conductor has a potential difference of 41 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{41}{5.0} = 8.2\;\Omega\)

Q153. A conductor has a potential difference of 30 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{30}{0.5} = 60.0\;\Omega\)

Q154. A conductor has a potential difference of 11 V across it and a current of 2.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{11}{2.5} = 4.4\;\Omega\)

Q155. A conductor has a potential difference of 4 V across it and a current of 4.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{4}{4.0} = 1.0\;\Omega\)

Q156. A conductor has a potential difference of 13 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{13}{0.25} = 52.0\;\Omega\)

Q157. A conductor has a potential difference of 8 V across it and a current of 0.1 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{8}{0.1} = 80.0\;\Omega\)

Q158. A conductor has a potential difference of 27 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{27}{3.0} = 9.0\;\Omega\)

Q159. A conductor has a potential difference of 47 V across it and a current of 1.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{47}{1.0} = 47.0\;\Omega\)

Q160. A conductor has a potential difference of 4 V across it and a current of 1.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{4}{1.5} = 2.67\;\Omega\)

Q161. A conductor has a potential difference of 33 V across it and a current of 1.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{33}{1.5} = 22.0\;\Omega\)

Q162. A conductor has a potential difference of 5 V across it and a current of 0.25 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{5}{0.25} = 20.0\;\Omega\)

Q163. A conductor has a potential difference of 22 V across it and a current of 5.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{22}{5.0} = 4.4\;\Omega\)

Q164. A conductor has a potential difference of 48 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{48}{3.0} = 16.0\;\Omega\)

Q165. A conductor has a potential difference of 35 V across it and a current of 2.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{35}{2.0} = 17.5\;\Omega\)

Q166. A conductor has a potential difference of 8 V across it and a current of 1.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{8}{1.0} = 8.0\;\Omega\)

Q167. A conductor has a potential difference of 45 V across it and a current of 3.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{45}{3.0} = 15.0\;\Omega\)

Q168. A conductor has a potential difference of 18 V across it and a current of 2.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{18}{2.5} = 7.2\;\Omega\)

Q169. A conductor has a potential difference of 38 V across it and a current of 1.0 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{38}{1.0} = 38.0\;\Omega\)

Q170. A conductor has a potential difference of 24 V across it and a current of 0.5 A flowing through it. Find its resistance.

Difficulty: Easy · Topic: Ohm's Law - Finding Resistance

Solution

By Ohm's Law: \(V = IR\)

\(R = \frac{V}{I} = \frac{24}{0.5} = 48.0\;\Omega\)

Q171. Two resistors \(R_1 = 13\;\Omega\) and \(R_2 = 11\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 13 + 11 = 24\;\Omega\)

Q172. Two resistors \(R_1 = 50\;\Omega\) and \(R_2 = 2\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 50 + 2 = 52\;\Omega\)

Q173. Two resistors \(R_1 = 27\;\Omega\) and \(R_2 = 15\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 27 + 15 = 42\;\Omega\)

Q174. Two resistors \(R_1 = 7\;\Omega\) and \(R_2 = 45\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 7 + 45 = 52\;\Omega\)

Q175. Two resistors \(R_1 = 26\;\Omega\) and \(R_2 = 23\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 26 + 23 = 49\;\Omega\)

Q176. Two resistors \(R_1 = 17\;\Omega\) and \(R_2 = 6\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 17 + 6 = 23\;\Omega\)

Q177. Two resistors \(R_1 = 5\;\Omega\) and \(R_2 = 48\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 5 + 48 = 53\;\Omega\)

Q178. Two resistors \(R_1 = 20\;\Omega\) and \(R_2 = 16\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 20 + 16 = 36\;\Omega\)

Q179. Two resistors \(R_1 = 28\;\Omega\) and \(R_2 = 46\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 28 + 46 = 74\;\Omega\)

Q180. Two resistors \(R_1 = 34\;\Omega\) and \(R_2 = 46\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 34 + 46 = 80\;\Omega\)

Q181. Two resistors \(R_1 = 13\;\Omega\) and \(R_2 = 29\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 13 + 29 = 42\;\Omega\)

Q182. Two resistors \(R_1 = 16\;\Omega\) and \(R_2 = 18\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 16 + 18 = 34\;\Omega\)

Q183. Two resistors \(R_1 = 16\;\Omega\) and \(R_2 = 2\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 16 + 2 = 18\;\Omega\)

Q184. Two resistors \(R_1 = 30\;\Omega\) and \(R_2 = 28\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 30 + 28 = 58\;\Omega\)

Q185. Two resistors \(R_1 = 47\;\Omega\) and \(R_2 = 4\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 47 + 4 = 51\;\Omega\)

Q186. Two resistors \(R_1 = 23\;\Omega\) and \(R_2 = 49\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 23 + 49 = 72\;\Omega\)

Q187. Two resistors \(R_1 = 42\;\Omega\) and \(R_2 = 38\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 42 + 38 = 80\;\Omega\)

Q188. Two resistors \(R_1 = 20\;\Omega\) and \(R_2 = 19\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 20 + 19 = 39\;\Omega\)

Q189. Two resistors \(R_1 = 10\;\Omega\) and \(R_2 = 26\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 10 + 26 = 36\;\Omega\)

Q190. Two resistors \(R_1 = 33\;\Omega\) and \(R_2 = 5\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 33 + 5 = 38\;\Omega\)

Q191. Two resistors \(R_1 = 28\;\Omega\) and \(R_2 = 34\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 28 + 34 = 62\;\Omega\)

Q192. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 40\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 25 + 40 = 65\;\Omega\)

Q193. Two resistors \(R_1 = 23\;\Omega\) and \(R_2 = 34\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 23 + 34 = 57\;\Omega\)

Q194. Two resistors \(R_1 = 3\;\Omega\) and \(R_2 = 23\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 3 + 23 = 26\;\Omega\)

Q195. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 36\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 25 + 36 = 61\;\Omega\)

Q196. Two resistors \(R_1 = 35\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 35 + 13 = 48\;\Omega\)

Q197. Two resistors \(R_1 = 14\;\Omega\) and \(R_2 = 23\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 14 + 23 = 37\;\Omega\)

Q198. Two resistors \(R_1 = 48\;\Omega\) and \(R_2 = 46\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 48 + 46 = 94\;\Omega\)

Q199. Two resistors \(R_1 = 48\;\Omega\) and \(R_2 = 35\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 48 + 35 = 83\;\Omega\)

Q200. Two resistors \(R_1 = 6\;\Omega\) and \(R_2 = 31\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 6 + 31 = 37\;\Omega\)

Q201. Two resistors \(R_1 = 12\;\Omega\) and \(R_2 = 10\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 12 + 10 = 22\;\Omega\)

Q202. Two resistors \(R_1 = 40\;\Omega\) and \(R_2 = 19\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 40 + 19 = 59\;\Omega\)

Q203. Two resistors \(R_1 = 5\;\Omega\) and \(R_2 = 29\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 5 + 29 = 34\;\Omega\)

Q204. Two resistors \(R_1 = 9\;\Omega\) and \(R_2 = 43\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 9 + 43 = 52\;\Omega\)

Q205. Two resistors \(R_1 = 19\;\Omega\) and \(R_2 = 18\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 19 + 18 = 37\;\Omega\)

Q206. Two resistors \(R_1 = 14\;\Omega\) and \(R_2 = 25\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 14 + 25 = 39\;\Omega\)

Q207. Two resistors \(R_1 = 5\;\Omega\) and \(R_2 = 41\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 5 + 41 = 46\;\Omega\)

Q208. Two resistors \(R_1 = 24\;\Omega\) and \(R_2 = 24\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 24 + 24 = 48\;\Omega\)

Q209. Two resistors \(R_1 = 22\;\Omega\) and \(R_2 = 37\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 22 + 37 = 59\;\Omega\)

Q210. Two resistors \(R_1 = 2\;\Omega\) and \(R_2 = 49\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 2 + 49 = 51\;\Omega\)

Q211. Two resistors \(R_1 = 32\;\Omega\) and \(R_2 = 29\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 32 + 29 = 61\;\Omega\)

Q212. Two resistors \(R_1 = 10\;\Omega\) and \(R_2 = 34\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 10 + 34 = 44\;\Omega\)

Q213. Two resistors \(R_1 = 37\;\Omega\) and \(R_2 = 23\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 37 + 23 = 60\;\Omega\)

Q214. Two resistors \(R_1 = 36\;\Omega\) and \(R_2 = 36\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 36 + 36 = 72\;\Omega\)

Q215. Two resistors \(R_1 = 36\;\Omega\) and \(R_2 = 38\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 36 + 38 = 74\;\Omega\)

Q216. Two resistors \(R_1 = 21\;\Omega\) and \(R_2 = 41\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 21 + 41 = 62\;\Omega\)

Q217. Two resistors \(R_1 = 42\;\Omega\) and \(R_2 = 30\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 42 + 30 = 72\;\Omega\)

Q218. Two resistors \(R_1 = 42\;\Omega\) and \(R_2 = 25\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 42 + 25 = 67\;\Omega\)

Q219. Two resistors \(R_1 = 38\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 38 + 13 = 51\;\Omega\)

Q220. Two resistors \(R_1 = 27\;\Omega\) and \(R_2 = 36\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 27 + 36 = 63\;\Omega\)

Q221. Two resistors \(R_1 = 30\;\Omega\) and \(R_2 = 35\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 30 + 35 = 65\;\Omega\)

Q222. Two resistors \(R_1 = 4\;\Omega\) and \(R_2 = 36\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 4 + 36 = 40\;\Omega\)

Q223. Two resistors \(R_1 = 49\;\Omega\) and \(R_2 = 2\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 49 + 2 = 51\;\Omega\)

Q224. Two resistors \(R_1 = 47\;\Omega\) and \(R_2 = 36\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 47 + 36 = 83\;\Omega\)

Q225. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 37\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 25 + 37 = 62\;\Omega\)

Q226. Two resistors \(R_1 = 48\;\Omega\) and \(R_2 = 32\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 48 + 32 = 80\;\Omega\)

Q227. Two resistors \(R_1 = 49\;\Omega\) and \(R_2 = 7\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 49 + 7 = 56\;\Omega\)

Q228. Two resistors \(R_1 = 30\;\Omega\) and \(R_2 = 7\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 30 + 7 = 37\;\Omega\)

Q229. Two resistors \(R_1 = 50\;\Omega\) and \(R_2 = 32\;\Omega\) are connected in series. Find the total resistance.

Difficulty: Easy · Topic: Resistors in Series

Solution

For series: \(R_{total} = R_1 + R_2\)

\(= 50 + 32 = 82\;\Omega\)

Q230. A current of 0.25 A flows through a resistance of 38 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 38 = 9.5\) V

Q231. A current of 2.0 A flows through a resistance of 78 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.0 \times 78 = 156.0\) V

Q232. A current of 0.25 A flows through a resistance of 83 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 83 = 20.75\) V

Q233. A current of 1.0 A flows through a resistance of 3 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 3 = 3.0\) V

Q234. A current of 0.2 A flows through a resistance of 72 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 72 = 14.4\) V

Q235. A current of 2.5 A flows through a resistance of 35 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.5 \times 35 = 87.5\) V

Q236. A current of 5.0 A flows through a resistance of 35 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 35 = 175.0\) V

Q237. A current of 4.0 A flows through a resistance of 25 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 4.0 \times 25 = 100.0\) V

Q238. A current of 1.0 A flows through a resistance of 22 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 22 = 22.0\) V

Q239. A current of 0.2 A flows through a resistance of 52 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 52 = 10.4\) V

Q240. A current of 3.0 A flows through a resistance of 90 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 3.0 \times 90 = 270.0\) V

Q241. A current of 1.0 A flows through a resistance of 75 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 75 = 75.0\) V

Q242. A current of 2.0 A flows through a resistance of 66 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.0 \times 66 = 132.0\) V

Q243. A current of 2.0 A flows through a resistance of 60 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.0 \times 60 = 120.0\) V

Q244. A current of 1.0 A flows through a resistance of 92 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 92 = 92.0\) V

Q245. A current of 2.5 A flows through a resistance of 34 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.5 \times 34 = 85.0\) V

Q246. A current of 0.2 A flows through a resistance of 6 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 6 = 1.2\) V

Q247. A current of 1.5 A flows through a resistance of 14 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 14 = 21.0\) V

Q248. A current of 5.0 A flows through a resistance of 83 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 83 = 415.0\) V

Q249. A current of 1.5 A flows through a resistance of 50 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 50 = 75.0\) V

Q250. A current of 2.5 A flows through a resistance of 96 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.5 \times 96 = 240.0\) V

Q251. A current of 4.0 A flows through a resistance of 99 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 4.0 \times 99 = 396.0\) V

Q252. A current of 0.25 A flows through a resistance of 30 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 30 = 7.5\) V

Q253. A current of 1.5 A flows through a resistance of 80 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 80 = 120.0\) V

Q254. A current of 5.0 A flows through a resistance of 84 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 84 = 420.0\) V

Q255. A current of 1.5 A flows through a resistance of 5 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 5 = 7.5\) V

Q256. A current of 5.0 A flows through a resistance of 86 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 86 = 430.0\) V

Q257. A current of 1.0 A flows through a resistance of 62 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 62 = 62.0\) V

Q258. A current of 1.5 A flows through a resistance of 46 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 46 = 69.0\) V

Q259. A current of 0.25 A flows through a resistance of 82 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 82 = 20.5\) V

Q260. A current of 0.2 A flows through a resistance of 14 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 14 = 2.8\) V

Q261. A current of 2.0 A flows through a resistance of 41 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.0 \times 41 = 82.0\) V

Q262. A current of 4.0 A flows through a resistance of 12 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 4.0 \times 12 = 48.0\) V

Q263. A current of 1.5 A flows through a resistance of 20 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 20 = 30.0\) V

Q264. A current of 2.5 A flows through a resistance of 17 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.5 \times 17 = 42.5\) V

Q265. A current of 1.0 A flows through a resistance of 7 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 7 = 7.0\) V

Q266. A current of 2.0 A flows through a resistance of 5 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.0 \times 5 = 10.0\) V

Q267. A current of 1.0 A flows through a resistance of 97 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.0 \times 97 = 97.0\) V

Q268. A current of 0.25 A flows through a resistance of 79 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 79 = 19.75\) V

Q269. A current of 0.25 A flows through a resistance of 21 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 21 = 5.25\) V

Q270. A current of 3.0 A flows through a resistance of 48 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 3.0 \times 48 = 144.0\) V

Q271. A current of 0.2 A flows through a resistance of 81 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 81 = 16.2\) V

Q272. A current of 0.25 A flows through a resistance of 3 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 3 = 0.75\) V

Q273. A current of 2.5 A flows through a resistance of 48 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.5 \times 48 = 120.0\) V

Q274. A current of 2.5 A flows through a resistance of 45 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.5 \times 45 = 112.5\) V

Q275. A current of 0.2 A flows through a resistance of 67 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 67 = 13.4\) V

Q276. A current of 0.2 A flows through a resistance of 39 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.2 \times 39 = 7.8\) V

Q277. A current of 2.0 A flows through a resistance of 15 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.0 \times 15 = 30.0\) V

Q278. A current of 2.5 A flows through a resistance of 63 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 2.5 \times 63 = 157.5\) V

Q279. A current of 0.5 A flows through a resistance of 28 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.5 \times 28 = 14.0\) V

Q280. A current of 5.0 A flows through a resistance of 44 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 5.0 \times 44 = 220.0\) V

Q281. A current of 1.5 A flows through a resistance of 68 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 1.5 \times 68 = 102.0\) V

Q282. A current of 0.25 A flows through a resistance of 9 ohm. Find the potential difference across the resistor.

Difficulty: Easy · Topic: Ohm's Law - Finding Voltage

Solution

By Ohm's Law: \(V = IR\)

\(V = 0.25 \times 9 = 2.25\) V

Q283. A 2000 W device runs for 3 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 3\text{ h} = 6.0\) kWh

Q284. A 750 W device runs for 10 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 750\text{ W} \times 10\text{ h} = 7.5\) kWh

Q285. A 500 W device runs for 7 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 500\text{ W} \times 7\text{ h} = 3.5\) kWh

Q286. A 500 W device runs for 4 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 500\text{ W} \times 4\text{ h} = 2.0\) kWh

Q287. A 2000 W device runs for 1 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 1\text{ h} = 2.0\) kWh

Q288. A 3000 W device runs for 12 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 12\text{ h} = 36.0\) kWh

Q289. A 750 W device runs for 9 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 750\text{ W} \times 9\text{ h} = 6.75\) kWh

Q290. A 500 W device runs for 5 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 500\text{ W} \times 5\text{ h} = 2.5\) kWh

Q291. A 100 W device runs for 8 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 100\text{ W} \times 8\text{ h} = 0.8\) kWh

Q292. A 200 W device runs for 10 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 200\text{ W} \times 10\text{ h} = 2.0\) kWh

Q293. A 1000 W device runs for 3 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 3\text{ h} = 3.0\) kWh

Q294. A 1000 W device runs for 12 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 12\text{ h} = 12.0\) kWh

Q295. A 1000 W device runs for 10 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 10\text{ h} = 10.0\) kWh

Q296. A 1000 W device runs for 5 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 5\text{ h} = 5.0\) kWh

Q297. A 2000 W device runs for 2 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 2\text{ h} = 4.0\) kWh

Q298. A 100 W device runs for 4 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 100\text{ W} \times 4\text{ h} = 0.4\) kWh

Q299. A 2000 W device runs for 10 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 10\text{ h} = 20.0\) kWh

Q300. A 200 W device runs for 11 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 200\text{ W} \times 11\text{ h} = 2.2\) kWh

Q301. A 1000 W device runs for 1 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 1\text{ h} = 1.0\) kWh

Q302. A 750 W device runs for 5 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 750\text{ W} \times 5\text{ h} = 3.75\) kWh

Q303. A 2000 W device runs for 11 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 11\text{ h} = 22.0\) kWh

Q304. A 500 W device runs for 9 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 500\text{ W} \times 9\text{ h} = 4.5\) kWh

Q305. A 3000 W device runs for 7 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 7\text{ h} = 21.0\) kWh

Q306. A 1000 W device runs for 7 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 7\text{ h} = 7.0\) kWh

Q307. A 1500 W device runs for 2 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 2\text{ h} = 3.0\) kWh

Q308. A 1500 W device runs for 4 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 4\text{ h} = 6.0\) kWh

Q309. A 3000 W device runs for 11 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 11\text{ h} = 33.0\) kWh

Q310. A 750 W device runs for 8 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 750\text{ W} \times 8\text{ h} = 6.0\) kWh

Q311. A 200 W device runs for 4 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 200\text{ W} \times 4\text{ h} = 0.8\) kWh

Q312. A 100 W device runs for 2 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 100\text{ W} \times 2\text{ h} = 0.2\) kWh

Q313. A 1000 W device runs for 6 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 6\text{ h} = 6.0\) kWh

Q314. A 100 W device runs for 6 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 100\text{ W} \times 6\text{ h} = 0.6\) kWh

Q315. A 100 W device runs for 12 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 100\text{ W} \times 12\text{ h} = 1.2\) kWh

Q316. A 100 W device runs for 3 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 100\text{ W} \times 3\text{ h} = 0.3\) kWh

Q317. A 2000 W device runs for 9 hours. Find the energy consumed in kWh.

Difficulty: Easy · Topic: Electrical Energy from Power and Time

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 9\text{ h} = 18.0\) kWh

Q318. A charge of 150 C flows through a wire in 1 minute. Calculate the electric current.

Difficulty: Easy-Medium · Topic: Electric Current

Solution

Given: Q = 150 C, t = 1 min = 60 s

Formula: I = Q/t

I = 150/60 = 2.5 A

Q319. How much work is done in moving a charge of 4 C across two points having a potential difference of 12 V?

Difficulty: Easy-Medium · Topic: Potential Difference

Solution

Given: Q = 4 C, V = 12 V

Formula: V = W/Q, so W = VQ

W = 12 × 4 = 48 J

48 joules of work is done (or 48 J of energy is transferred) in moving 4 C of charge across a potential difference of 12 V.

Q320. The V-I graph for an ohmic conductor is:

Difficulty: Easy-Medium · Topic: Ohm's Law

Solution

For an ohmic conductor (one that obeys Ohm's law), V is directly proportional to I at constant temperature. The graph of V vs I is a straight line passing through the origin. The slope of this line equals the resistance R. If the line is steeper, the resistance is higher.

Q321. A conductor has a resistance of 20 Ω. If a potential difference of 100 V is applied across it, find the current flowing through it.

Difficulty: Easy-Medium · Topic: Ohm's Law

Solution

Given: R = 20 Ω, V = 100 V

Using Ohm's law: V = IR

I = V/R = 100/20 = 5 A

Q322. State Ohm's law. What are the limitations of Ohm's law?

Difficulty: Easy-Medium · Topic: Ohm's Law

Solution

Ohm's Law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference applied across its ends.

V ∝ I (at constant temperature)

V = IR, where R is the resistance of the conductor.

Limitations:

  • Ohm's law is not valid for non-ohmic conductors like diodes, transistors, and electrolytes.
  • It does not hold for semiconductor devices where the V-I relationship is non-linear.
  • It is not applicable when temperature changes significantly (since R changes with temperature for most materials).
  • It does not apply to devices where current depends on the polarity of voltage (e.g., LED, diode).
Q323. Alloys are commonly used in electrical heating devices because:

Difficulty: Easy-Medium · Topic: Resistivity

Solution

Alloys like nichrome and constantan are used in heating devices because they have:

  • High resistivity — produces more heat for the same current (H = I²Rt, higher R means more heat).
  • High melting point — they can withstand the high temperatures produced without melting.
  • Resistance does not change much with temperature — ensures consistent performance.
Q324. In a series circuit, if one component fails (breaks), what happens?

Difficulty: Easy-Medium · Topic: Resistors in Series

Solution

In a series circuit, there is only one path for current to flow. If any component breaks, the circuit is broken and current stops flowing through all components. This is a disadvantage of series connections. In parallel circuits, each component has its own path, so other components continue working even if one fails.

Q325. 1 kWh is equal to:

Difficulty: Easy-Medium · Topic: Electric Power

Solution

1 kWh = 1 kilowatt × 1 hour

= 1000 W × 3600 s

= 1000 × 3600 J

= 3.6 × 10⁶ J = 3.6 MJ

Q326. What is an electric fuse? Explain its working principle.

Difficulty: Easy-Medium · Topic: Heating Effect

Solution

An electric fuse is a safety device that protects electrical circuits and appliances from excessive current (overcurrent) or short circuits.

Construction: It consists of a short piece of wire made of a material with a low melting point (such as tin or an alloy of tin and lead). The fuse wire is connected in series with the circuit.

Working principle: The fuse works on the heating effect of electric current (Joule heating, H = I²Rt).

  • Under normal conditions, the current through the fuse wire is within its rated value, and the heat produced is safely dissipated.
  • If the current exceeds the rated value (due to a short circuit or overload), the heat produced (I²Rt) becomes very large.
  • This excessive heat melts the fuse wire, breaking the circuit and stopping current flow.
  • This protects the wiring and appliances from damage or fire.
Q327. An ammeter is connected in a circuit. It should be connected:

Difficulty: Easy-Medium · Topic: Electric Current

Solution

An ammeter measures current and must be connected in series so that all the current flowing through the circuit passes through it. It has very low resistance to avoid significantly affecting the circuit. A voltmeter, on the other hand, is connected in parallel. Connecting an ammeter in parallel would create a short circuit (due to its very low resistance) and could damage the instrument.

Q328. Two resistors \(R_1 = 8\;\Omega\) and \(R_2 = 3\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 3}{8+3} = 2.18\;\Omega\)

Q329. Two resistors \(R_1 = 24\;\Omega\) and \(R_2 = 28\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{24 \times 28}{24+28} = 12.92\;\Omega\)

Q330. Two resistors \(R_1 = 26\;\Omega\) and \(R_2 = 42\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{26 \times 42}{26+42} = 16.06\;\Omega\)

Q331. Two resistors \(R_1 = 33\;\Omega\) and \(R_2 = 7\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{33 \times 7}{33+7} = 5.78\;\Omega\)

Q332. Two resistors \(R_1 = 22\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{22 \times 13}{22+13} = 8.17\;\Omega\)

Q333. Two resistors \(R_1 = 8\;\Omega\) and \(R_2 = 25\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 25}{8+25} = 6.06\;\Omega\)

Q334. Two resistors \(R_1 = 27\;\Omega\) and \(R_2 = 21\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{27 \times 21}{27+21} = 11.81\;\Omega\)

Q335. Two resistors \(R_1 = 28\;\Omega\) and \(R_2 = 28\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{28 \times 28}{28+28} = 14.0\;\Omega\)

Q336. Two resistors \(R_1 = 10\;\Omega\) and \(R_2 = 34\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \times 34}{10+34} = 7.73\;\Omega\)

Q337. Two resistors \(R_1 = 18\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{18 \times 13}{18+13} = 7.55\;\Omega\)

Q338. Two resistors \(R_1 = 20\;\Omega\) and \(R_2 = 6\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{20 \times 6}{20+6} = 4.62\;\Omega\)

Q339. Two resistors \(R_1 = 32\;\Omega\) and \(R_2 = 49\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{32 \times 49}{32+49} = 19.36\;\Omega\)

Q340. Two resistors \(R_1 = 47\;\Omega\) and \(R_2 = 37\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{47 \times 37}{47+37} = 20.7\;\Omega\)

Q341. Two resistors \(R_1 = 7\;\Omega\) and \(R_2 = 24\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{7 \times 24}{7+24} = 5.42\;\Omega\)

Q342. Two resistors \(R_1 = 2\;\Omega\) and \(R_2 = 10\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{2 \times 10}{2+10} = 1.67\;\Omega\)

Q343. Two resistors \(R_1 = 48\;\Omega\) and \(R_2 = 32\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{48 \times 32}{48+32} = 19.2\;\Omega\)

Q344. Two resistors \(R_1 = 50\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{50 \times 13}{50+13} = 10.32\;\Omega\)

Q345. Two resistors \(R_1 = 41\;\Omega\) and \(R_2 = 44\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{41 \times 44}{41+44} = 21.22\;\Omega\)

Q346. Two resistors \(R_1 = 38\;\Omega\) and \(R_2 = 37\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{38 \times 37}{38+37} = 18.75\;\Omega\)

Q347. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 10\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{25 \times 10}{25+10} = 7.14\;\Omega\)

Q348. Two resistors \(R_1 = 23\;\Omega\) and \(R_2 = 4\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{23 \times 4}{23+4} = 3.41\;\Omega\)

Q349. Two resistors \(R_1 = 10\;\Omega\) and \(R_2 = 2\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \times 2}{10+2} = 1.67\;\Omega\)

Q350. Two resistors \(R_1 = 2\;\Omega\) and \(R_2 = 4\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{2 \times 4}{2+4} = 1.33\;\Omega\)

Q351. Two resistors \(R_1 = 12\;\Omega\) and \(R_2 = 9\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{12 \times 9}{12+9} = 5.14\;\Omega\)

Q352. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 50\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{25 \times 50}{25+50} = 16.67\;\Omega\)

Q353. Two resistors \(R_1 = 23\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{23 \times 13}{23+13} = 8.31\;\Omega\)

Q354. Two resistors \(R_1 = 50\;\Omega\) and \(R_2 = 31\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{50 \times 31}{50+31} = 19.14\;\Omega\)

Q355. Two resistors \(R_1 = 24\;\Omega\) and \(R_2 = 9\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{24 \times 9}{24+9} = 6.55\;\Omega\)

Q356. Two resistors \(R_1 = 33\;\Omega\) and \(R_2 = 12\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{33 \times 12}{33+12} = 8.8\;\Omega\)

Q357. Two resistors \(R_1 = 12\;\Omega\) and \(R_2 = 3\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{12 \times 3}{12+3} = 2.4\;\Omega\)

Q358. An appliance is rated 500 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{500}{220} = 2.27\) A

Q359. An appliance is rated 2000 W at 240 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{2000}{240} = 8.33\) A

Q360. An appliance is rated 500 W at 240 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{500}{240} = 2.08\) A

Q361. An appliance is rated 750 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{750}{230} = 3.26\) A

Q362. An appliance is rated 200 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{200}{220} = 0.91\) A

Q363. An appliance is rated 200 W at 240 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{200}{240} = 0.83\) A

Q364. An appliance is rated 100 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{100}{230} = 0.43\) A

Q365. An appliance is rated 750 W at 110 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{750}{110} = 6.82\) A

Q366. An appliance is rated 60 W at 110 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{60}{110} = 0.55\) A

Q367. An appliance is rated 200 W at 110 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{200}{110} = 1.82\) A

Q368. An appliance is rated 2000 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{2000}{230} = 8.7\) A

Q369. An appliance is rated 1000 W at 110 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{1000}{110} = 9.09\) A

Q370. An appliance is rated 1500 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{1500}{230} = 6.52\) A

Q371. An appliance is rated 100 W at 240 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{100}{240} = 0.42\) A

Q372. An appliance is rated 40 W at 240 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{40}{240} = 0.17\) A

Q373. An appliance is rated 40 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{40}{220} = 0.18\) A

Q374. An appliance is rated 1000 W at 240 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{1000}{240} = 4.17\) A

Q375. An appliance is rated 1500 W at 240 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{1500}{240} = 6.25\) A

Q376. An appliance is rated 1000 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{1000}{230} = 4.35\) A

Q377. An appliance is rated 1500 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{1500}{220} = 6.82\) A

Q378. An appliance is rated 60 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{60}{220} = 0.27\) A

Q379. An appliance is rated 1000 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{1000}{220} = 4.55\) A

Q380. A 2000 W heater is used for 8 hours daily for 20 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 8\text{ h/day} \times 20\text{ days}\)

\(= 2000 \times 8 \times 20 = \) ... Wh

In kWh: divide by 1000 \(= 320.0\) kWh

Q381. A 500 W heater is used for 2 hours daily for 20 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 2\text{ h/day} \times 20\text{ days}\)

\(= 500 \times 2 \times 20 = \) ... Wh

In kWh: divide by 1000 \(= 20.0\) kWh

Q382. A 3000 W heater is used for 5 hours daily for 21 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 5\text{ h/day} \times 21\text{ days}\)

\(= 3000 \times 5 \times 21 = \) ... Wh

In kWh: divide by 1000 \(= 315.0\) kWh

Q383. A 3000 W heater is used for 3 hours daily for 7 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 3\text{ h/day} \times 7\text{ days}\)

\(= 3000 \times 3 \times 7 = \) ... Wh

In kWh: divide by 1000 \(= 63.0\) kWh

Q384. A 1000 W heater is used for 8 hours daily for 6 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 8\text{ h/day} \times 6\text{ days}\)

\(= 1000 \times 8 \times 6 = \) ... Wh

In kWh: divide by 1000 \(= 48.0\) kWh

Q385. A 3000 W heater is used for 7 hours daily for 23 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 7\text{ h/day} \times 23\text{ days}\)

\(= 3000 \times 7 \times 23 = \) ... Wh

In kWh: divide by 1000 \(= 483.0\) kWh

Q386. A 3000 W heater is used for 4 hours daily for 8 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 4\text{ h/day} \times 8\text{ days}\)

\(= 3000 \times 4 \times 8 = \) ... Wh

In kWh: divide by 1000 \(= 96.0\) kWh

Q387. A 1000 W heater is used for 6 hours daily for 9 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 6\text{ h/day} \times 9\text{ days}\)

\(= 1000 \times 6 \times 9 = \) ... Wh

In kWh: divide by 1000 \(= 54.0\) kWh

Q388. A 1000 W heater is used for 4 hours daily for 11 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 4\text{ h/day} \times 11\text{ days}\)

\(= 1000 \times 4 \times 11 = \) ... Wh

In kWh: divide by 1000 \(= 44.0\) kWh

Q389. A 3000 W heater is used for 5 hours daily for 25 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 5\text{ h/day} \times 25\text{ days}\)

\(= 3000 \times 5 \times 25 = \) ... Wh

In kWh: divide by 1000 \(= 375.0\) kWh

Q390. A 1000 W heater is used for 5 hours daily for 6 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 5\text{ h/day} \times 6\text{ days}\)

\(= 1000 \times 5 \times 6 = \) ... Wh

In kWh: divide by 1000 \(= 30.0\) kWh

Q391. A 3000 W heater is used for 8 hours daily for 6 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 8\text{ h/day} \times 6\text{ days}\)

\(= 3000 \times 8 \times 6 = \) ... Wh

In kWh: divide by 1000 \(= 144.0\) kWh

Q392. A 3000 W heater is used for 1 hours daily for 15 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 1\text{ h/day} \times 15\text{ days}\)

\(= 3000 \times 1 \times 15 = \) ... Wh

In kWh: divide by 1000 \(= 45.0\) kWh

Q393. A 1500 W heater is used for 2 hours daily for 9 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 2\text{ h/day} \times 9\text{ days}\)

\(= 1500 \times 2 \times 9 = \) ... Wh

In kWh: divide by 1000 \(= 27.0\) kWh

Q394. A 3000 W heater is used for 2 hours daily for 19 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 2\text{ h/day} \times 19\text{ days}\)

\(= 3000 \times 2 \times 19 = \) ... Wh

In kWh: divide by 1000 \(= 114.0\) kWh

Q395. A 1500 W heater is used for 6 hours daily for 18 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 6\text{ h/day} \times 18\text{ days}\)

\(= 1500 \times 6 \times 18 = \) ... Wh

In kWh: divide by 1000 \(= 162.0\) kWh

Q396. A 2500 W heater is used for 3 hours daily for 21 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2500\text{ W} \times 3\text{ h/day} \times 21\text{ days}\)

\(= 2500 \times 3 \times 21 = \) ... Wh

In kWh: divide by 1000 \(= 157.5\) kWh

Q397. A 1000 W heater is used for 6 hours daily for 23 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 6\text{ h/day} \times 23\text{ days}\)

\(= 1000 \times 6 \times 23 = \) ... Wh

In kWh: divide by 1000 \(= 138.0\) kWh

Q398. A 1000 W heater is used for 5 hours daily for 19 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 5\text{ h/day} \times 19\text{ days}\)

\(= 1000 \times 5 \times 19 = \) ... Wh

In kWh: divide by 1000 \(= 95.0\) kWh

Q399. A 500 W heater is used for 7 hours daily for 17 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 7\text{ h/day} \times 17\text{ days}\)

\(= 500 \times 7 \times 17 = \) ... Wh

In kWh: divide by 1000 \(= 59.5\) kWh

Q400. A 2500 W heater is used for 2 hours daily for 10 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2500\text{ W} \times 2\text{ h/day} \times 10\text{ days}\)

\(= 2500 \times 2 \times 10 = \) ... Wh

In kWh: divide by 1000 \(= 50.0\) kWh

Q401. A 3000 W heater is used for 2 hours daily for 16 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 2\text{ h/day} \times 16\text{ days}\)

\(= 3000 \times 2 \times 16 = \) ... Wh

In kWh: divide by 1000 \(= 96.0\) kWh

Q402. A 500 W heater is used for 4 hours daily for 14 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 4\text{ h/day} \times 14\text{ days}\)

\(= 500 \times 4 \times 14 = \) ... Wh

In kWh: divide by 1000 \(= 28.0\) kWh

Q403. A 1500 W heater is used for 6 hours daily for 28 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 6\text{ h/day} \times 28\text{ days}\)

\(= 1500 \times 6 \times 28 = \) ... Wh

In kWh: divide by 1000 \(= 252.0\) kWh

Q404. A 2000 W heater is used for 7 hours daily for 11 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 7\text{ h/day} \times 11\text{ days}\)

\(= 2000 \times 7 \times 11 = \) ... Wh

In kWh: divide by 1000 \(= 154.0\) kWh

Q405. A 2500 W heater is used for 4 hours daily for 8 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2500\text{ W} \times 4\text{ h/day} \times 8\text{ days}\)

\(= 2500 \times 4 \times 8 = \) ... Wh

In kWh: divide by 1000 \(= 80.0\) kWh

Q406. A 1000 W heater is used for 2 hours daily for 10 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 2\text{ h/day} \times 10\text{ days}\)

\(= 1000 \times 2 \times 10 = \) ... Wh

In kWh: divide by 1000 \(= 20.0\) kWh

Q407. A 3000 W heater is used for 1 hours daily for 8 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 1\text{ h/day} \times 8\text{ days}\)

\(= 3000 \times 1 \times 8 = \) ... Wh

In kWh: divide by 1000 \(= 24.0\) kWh

Q408. A 500 W heater is used for 5 hours daily for 28 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 5\text{ h/day} \times 28\text{ days}\)

\(= 500 \times 5 \times 28 = \) ... Wh

In kWh: divide by 1000 \(= 70.0\) kWh

Q409. A current of 4.0 A flows through a resistance of 38 ohm for 27 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 38 \times 27 = 16416.0\) J

Q410. A current of 1.0 A flows through a resistance of 28 ohm for 64 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 28 \times 64 = 1792.0\) J

Q411. A current of 4.0 A flows through a resistance of 24 ohm for 35 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 24 \times 35 = 13440.0\) J

Q412. A current of 5.0 A flows through a resistance of 13 ohm for 87 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 13 \times 87 = 28275.0\) J

Q413. A current of 4.0 A flows through a resistance of 44 ohm for 32 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 44 \times 32 = 22528.0\) J

Q414. A current of 3.0 A flows through a resistance of 14 ohm for 36 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 14 \times 36 = 4536.0\) J

Q415. A current of 1.0 A flows through a resistance of 23 ohm for 19 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 23 \times 19 = 437.0\) J

Q416. A current of 2.0 A flows through a resistance of 8 ohm for 39 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 8 \times 39 = 1248.0\) J

Q417. A current of 3.0 A flows through a resistance of 22 ohm for 53 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 22 \times 53 = 10494.0\) J

Q418. A current of 5.0 A flows through a resistance of 42 ohm for 77 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 42 \times 77 = 80850.0\) J

Q419. A current of 5.0 A flows through a resistance of 22 ohm for 66 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 22 \times 66 = 36300.0\) J

Q420. A current of 0.5 A flows through a resistance of 47 ohm for 13 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 47 \times 13 = 152.75\) J

Q421. A current of 1.0 A flows through a resistance of 40 ohm for 34 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 40 \times 34 = 1360.0\) J

Q422. A current of 0.5 A flows through a resistance of 13 ohm for 18 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 13 \times 18 = 58.5\) J

Q423. A current of 2.0 A flows through a resistance of 45 ohm for 74 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 45 \times 74 = 13320.0\) J

Q424. A current of 2.0 A flows through a resistance of 41 ohm for 55 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 41 \times 55 = 9020.0\) J

Q425. A current of 1.0 A flows through a resistance of 38 ohm for 119 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 38 \times 119 = 4522.0\) J

Q426. A current of 0.5 A flows through a resistance of 32 ohm for 23 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 32 \times 23 = 184.0\) J

Q427. A current of 2.0 A flows through a resistance of 28 ohm for 96 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 28 \times 96 = 10752.0\) J

Q428. A current of 0.5 A flows through a resistance of 38 ohm for 59 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 38 \times 59 = 560.5\) J

Q429. A current of 1.0 A flows through a resistance of 7 ohm for 78 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 7 \times 78 = 546.0\) J

Q430. A current of 4.0 A flows through a resistance of 36 ohm for 110 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 36 \times 110 = 63360.0\) J

Q431. A current of 5.0 A flows through a resistance of 15 ohm for 46 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 15 \times 46 = 17250.0\) J

Q432. A current of 5.0 A flows through a resistance of 6 ohm for 106 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 6 \times 106 = 15900.0\) J

Q433. A current of 5.0 A flows through a resistance of 40 ohm for 18 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 40 \times 18 = 18000.0\) J

Q434. A current of 5.0 A flows through a resistance of 23 ohm for 53 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 23 \times 53 = 30475.0\) J

Q435. A current of 5.0 A flows through a resistance of 9 ohm for 99 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 9 \times 99 = 22275.0\) J

Q436. A current of 2.0 A flows through a resistance of 42 ohm for 64 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 42 \times 64 = 10752.0\) J

Q437. A current of 3.0 A flows through a resistance of 6 ohm for 49 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 6 \times 49 = 2646.0\) J

Q438. A current of 5.0 A flows through a resistance of 33 ohm for 64 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 33 \times 64 = 52800.0\) J

Q439. A wire of resistance 35 ohm has length 2 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{35 \times 0.1 \times 10^{-6}}{2} = 1.75e-06\) ohm.m

Q440. A wire of resistance 41 ohm has length 5 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{41 \times 0.25 \times 10^{-6}}{5} = 2.05e-06\) ohm.m

Q441. A wire of resistance 11 ohm has length 3 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{11 \times 0.5 \times 10^{-6}}{3} = 1.83e-06\) ohm.m

Q442. A wire of resistance 10 ohm has length 0.5 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{10 \times 2.0 \times 10^{-6}}{0.5} = 4.00e-05\) ohm.m

Q443. A wire of resistance 35 ohm has length 2 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{35 \times 1.0 \times 10^{-6}}{2} = 1.75e-05\) ohm.m

Q444. A wire of resistance 8 ohm has length 5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{8 \times 1.5 \times 10^{-6}}{5} = 2.40e-06\) ohm.m

Q445. A wire of resistance 5 ohm has length 2 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{5 \times 0.25 \times 10^{-6}}{2} = 6.25e-07\) ohm.m

Q446. A wire of resistance 11 ohm has length 1 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{11 \times 1.5 \times 10^{-6}}{1} = 1.65e-05\) ohm.m

Q447. A wire of resistance 41 ohm has length 0.5 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{41 \times 0.25 \times 10^{-6}}{0.5} = 2.05e-05\) ohm.m

Q448. A wire of resistance 48 ohm has length 0.5 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{48 \times 2.0 \times 10^{-6}}{0.5} = 1.92e-04\) ohm.m

Q449. A wire of resistance 34 ohm has length 5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{34 \times 1.5 \times 10^{-6}}{5} = 1.02e-05\) ohm.m

Q450. A wire of resistance 4 ohm has length 10 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{4 \times 1.0 \times 10^{-6}}{10} = 4.00e-07\) ohm.m

Q451. A wire of resistance 30 ohm has length 1 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{30 \times 0.25 \times 10^{-6}}{1} = 7.50e-06\) ohm.m

Q452. A wire of resistance 24 ohm has length 1 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{24 \times 1.5 \times 10^{-6}}{1} = 3.60e-05\) ohm.m

Q453. A wire of resistance 48 ohm has length 5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{48 \times 1.5 \times 10^{-6}}{5} = 1.44e-05\) ohm.m

Q454. A wire of resistance 29 ohm has length 5 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{29 \times 0.25 \times 10^{-6}}{5} = 1.45e-06\) ohm.m

Q455. A wire of resistance 47 ohm has length 1 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{47 \times 2.0 \times 10^{-6}}{1} = 9.40e-05\) ohm.m

Q456. A wire of resistance 36 ohm has length 5 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{36 \times 0.5 \times 10^{-6}}{5} = 3.60e-06\) ohm.m

Q457. A wire of resistance 18 ohm has length 3 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{18 \times 0.1 \times 10^{-6}}{3} = 6.00e-07\) ohm.m

Q458. A wire of resistance 33 ohm has length 10 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{33 \times 0.1 \times 10^{-6}}{10} = 3.30e-07\) ohm.m

Q459. A wire of resistance 36 ohm has length 1 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{36 \times 2.0 \times 10^{-6}}{1} = 7.20e-05\) ohm.m

Q460. A wire of resistance 31 ohm has length 0.5 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{31 \times 2.0 \times 10^{-6}}{0.5} = 1.24e-04\) ohm.m

Q461. A wire of resistance 23 ohm has length 10 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{23 \times 0.5 \times 10^{-6}}{10} = 1.15e-06\) ohm.m

Q462. A wire of resistance 36 ohm has length 3 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{36 \times 1.0 \times 10^{-6}}{3} = 1.20e-05\) ohm.m

Q463. A wire of resistance 22 ohm has length 5 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{22 \times 1.0 \times 10^{-6}}{5} = 4.40e-06\) ohm.m

Q464. A wire of resistance 37 ohm has length 0.5 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{37 \times 0.1 \times 10^{-6}}{0.5} = 7.40e-06\) ohm.m

Q465. A wire of resistance 20 ohm has length 2 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{20 \times 2.0 \times 10^{-6}}{2} = 2.00e-05\) ohm.m

Q466. A wire of resistance 46 ohm has length 3 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{46 \times 2.0 \times 10^{-6}}{3} = 3.07e-05\) ohm.m

Q467. A wire of resistance 13 ohm has length 2 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{13 \times 2.0 \times 10^{-6}}{2} = 1.30e-05\) ohm.m

Q468. A wire of resistance 24 ohm has length 3 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{24 \times 1.5 \times 10^{-6}}{3} = 1.20e-05\) ohm.m

Q469. Two resistors \(R_1 = 49\;\Omega\) and \(R_2 = 37\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{49 \times 37}{49+37} = 21.08\;\Omega\)

Q470. Two resistors \(R_1 = 16\;\Omega\) and \(R_2 = 27\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{16 \times 27}{16+27} = 10.05\;\Omega\)

Q471. Two resistors \(R_1 = 39\;\Omega\) and \(R_2 = 24\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{39 \times 24}{39+24} = 14.86\;\Omega\)

Q472. Two resistors \(R_1 = 9\;\Omega\) and \(R_2 = 15\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{9 \times 15}{9+15} = 5.62\;\Omega\)

Q473. Two resistors \(R_1 = 18\;\Omega\) and \(R_2 = 38\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{18 \times 38}{18+38} = 12.21\;\Omega\)

Q474. Two resistors \(R_1 = 36\;\Omega\) and \(R_2 = 45\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{36 \times 45}{36+45} = 20.0\;\Omega\)

Q475. Two resistors \(R_1 = 34\;\Omega\) and \(R_2 = 29\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{34 \times 29}{34+29} = 15.65\;\Omega\)

Q476. Two resistors \(R_1 = 18\;\Omega\) and \(R_2 = 11\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{18 \times 11}{18+11} = 6.83\;\Omega\)

Q477. Two resistors \(R_1 = 8\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 13}{8+13} = 4.95\;\Omega\)

Q478. Two resistors \(R_1 = 27\;\Omega\) and \(R_2 = 2\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{27 \times 2}{27+2} = 1.86\;\Omega\)

Q479. Two resistors \(R_1 = 31\;\Omega\) and \(R_2 = 27\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{31 \times 27}{31+27} = 14.43\;\Omega\)

Q480. Two resistors \(R_1 = 41\;\Omega\) and \(R_2 = 20\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{41 \times 20}{41+20} = 13.44\;\Omega\)

Q481. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 41\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{25 \times 41}{25+41} = 15.53\;\Omega\)

Q482. Two resistors \(R_1 = 6\;\Omega\) and \(R_2 = 5\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6 \times 5}{6+5} = 2.73\;\Omega\)

Q483. Two resistors \(R_1 = 39\;\Omega\) and \(R_2 = 9\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{39 \times 9}{39+9} = 7.31\;\Omega\)

Q484. Two resistors \(R_1 = 50\;\Omega\) and \(R_2 = 30\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{50 \times 30}{50+30} = 18.75\;\Omega\)

Q485. Two resistors \(R_1 = 7\;\Omega\) and \(R_2 = 5\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{7 \times 5}{7+5} = 2.92\;\Omega\)

Q486. Two resistors \(R_1 = 40\;\Omega\) and \(R_2 = 20\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{40 \times 20}{40+20} = 13.33\;\Omega\)

Q487. Two resistors \(R_1 = 22\;\Omega\) and \(R_2 = 48\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{22 \times 48}{22+48} = 15.09\;\Omega\)

Q488. Two resistors \(R_1 = 38\;\Omega\) and \(R_2 = 18\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{38 \times 18}{38+18} = 12.21\;\Omega\)

Q489. Two resistors \(R_1 = 29\;\Omega\) and \(R_2 = 31\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{29 \times 31}{29+31} = 14.98\;\Omega\)

Q490. Two resistors \(R_1 = 21\;\Omega\) and \(R_2 = 33\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{21 \times 33}{21+33} = 12.83\;\Omega\)

Q491. Two resistors \(R_1 = 5\;\Omega\) and \(R_2 = 40\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{5 \times 40}{5+40} = 4.44\;\Omega\)

Q492. Two resistors \(R_1 = 8\;\Omega\) and \(R_2 = 36\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 36}{8+36} = 6.55\;\Omega\)

Q493. Two resistors \(R_1 = 23\;\Omega\) and \(R_2 = 45\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{23 \times 45}{23+45} = 15.22\;\Omega\)

Q494. Two resistors \(R_1 = 37\;\Omega\) and \(R_2 = 27\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{37 \times 27}{37+27} = 15.61\;\Omega\)

Q495. Two resistors \(R_1 = 36\;\Omega\) and \(R_2 = 42\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{36 \times 42}{36+42} = 19.38\;\Omega\)

Q496. Two resistors \(R_1 = 46\;\Omega\) and \(R_2 = 38\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{46 \times 38}{46+38} = 20.81\;\Omega\)

Q497. Two resistors \(R_1 = 48\;\Omega\) and \(R_2 = 16\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{48 \times 16}{48+16} = 12.0\;\Omega\)

Q498. Two resistors \(R_1 = 3\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{3 \times 13}{3+13} = 2.44\;\Omega\)

Q499. Two resistors \(R_1 = 28\;\Omega\) and \(R_2 = 49\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{28 \times 49}{28+49} = 17.82\;\Omega\)

Q500. Two resistors \(R_1 = 18\;\Omega\) and \(R_2 = 36\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{18 \times 36}{18+36} = 12.0\;\Omega\)

Q501. Two resistors \(R_1 = 45\;\Omega\) and \(R_2 = 49\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{45 \times 49}{45+49} = 23.46\;\Omega\)

Q502. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 8\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{25 \times 8}{25+8} = 6.06\;\Omega\)

Q503. Two resistors \(R_1 = 22\;\Omega\) and \(R_2 = 49\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{22 \times 49}{22+49} = 15.18\;\Omega\)

Q504. Two resistors \(R_1 = 34\;\Omega\) and \(R_2 = 6\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{34 \times 6}{34+6} = 5.1\;\Omega\)

Q505. Two resistors \(R_1 = 32\;\Omega\) and \(R_2 = 42\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{32 \times 42}{32+42} = 18.16\;\Omega\)

Q506. Two resistors \(R_1 = 37\;\Omega\) and \(R_2 = 24\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{37 \times 24}{37+24} = 14.56\;\Omega\)

Q507. Two resistors \(R_1 = 37\;\Omega\) and \(R_2 = 42\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{37 \times 42}{37+42} = 19.67\;\Omega\)

Q508. Two resistors \(R_1 = 35\;\Omega\) and \(R_2 = 44\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{35 \times 44}{35+44} = 19.49\;\Omega\)

Q509. Two resistors \(R_1 = 14\;\Omega\) and \(R_2 = 18\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{14 \times 18}{14+18} = 7.88\;\Omega\)

Q510. Two resistors \(R_1 = 9\;\Omega\) and \(R_2 = 39\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{9 \times 39}{9+39} = 7.31\;\Omega\)

Q511. Two resistors \(R_1 = 25\;\Omega\) and \(R_2 = 38\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{25 \times 38}{25+38} = 15.08\;\Omega\)

Q512. Two resistors \(R_1 = 37\;\Omega\) and \(R_2 = 45\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{37 \times 45}{37+45} = 20.3\;\Omega\)

Q513. Two resistors \(R_1 = 40\;\Omega\) and \(R_2 = 10\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{40 \times 10}{40+10} = 8.0\;\Omega\)

Q514. Two resistors \(R_1 = 14\;\Omega\) and \(R_2 = 21\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{14 \times 21}{14+21} = 8.4\;\Omega\)

Q515. Two resistors \(R_1 = 47\;\Omega\) and \(R_2 = 29\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{47 \times 29}{47+29} = 17.93\;\Omega\)

Q516. Two resistors \(R_1 = 23\;\Omega\) and \(R_2 = 26\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{23 \times 26}{23+26} = 12.2\;\Omega\)

Q517. Two resistors \(R_1 = 43\;\Omega\) and \(R_2 = 23\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{43 \times 23}{43+23} = 14.98\;\Omega\)

Q518. Two resistors \(R_1 = 40\;\Omega\) and \(R_2 = 16\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{40 \times 16}{40+16} = 11.43\;\Omega\)

Q519. Two resistors \(R_1 = 23\;\Omega\) and \(R_2 = 35\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{23 \times 35}{23+35} = 13.88\;\Omega\)

Q520. Two resistors \(R_1 = 33\;\Omega\) and \(R_2 = 3\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{33 \times 3}{33+3} = 2.75\;\Omega\)

Q521. Two resistors \(R_1 = 29\;\Omega\) and \(R_2 = 15\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{29 \times 15}{29+15} = 9.89\;\Omega\)

Q522. Two resistors \(R_1 = 4\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{4 \times 13}{4+13} = 3.06\;\Omega\)

Q523. Two resistors \(R_1 = 39\;\Omega\) and \(R_2 = 12\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{39 \times 12}{39+12} = 9.18\;\Omega\)

Q524. Two resistors \(R_1 = 34\;\Omega\) and \(R_2 = 46\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{34 \times 46}{34+46} = 19.55\;\Omega\)

Q525. Two resistors \(R_1 = 43\;\Omega\) and \(R_2 = 37\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{43 \times 37}{43+37} = 19.89\;\Omega\)

Q526. Two resistors \(R_1 = 35\;\Omega\) and \(R_2 = 41\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{35 \times 41}{35+41} = 18.88\;\Omega\)

Q527. Two resistors \(R_1 = 30\;\Omega\) and \(R_2 = 13\;\Omega\) are connected in parallel. Find the equivalent resistance.

Difficulty: Easy-Medium · Topic: Resistors in Parallel

Solution

For parallel: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{30 \times 13}{30+13} = 9.07\;\Omega\)

Q528. An appliance is rated 500 W at 110 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{500}{110} = 4.55\) A

Q529. An appliance is rated 750 W at 240 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{750}{240} = 3.12\) A

Q530. An appliance is rated 40 W at 110 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{40}{110} = 0.36\) A

Q531. An appliance is rated 200 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{200}{230} = 0.87\) A

Q532. An appliance is rated 750 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{750}{220} = 3.41\) A

Q533. An appliance is rated 40 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{40}{230} = 0.17\) A

Q534. An appliance is rated 60 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{60}{230} = 0.26\) A

Q535. An appliance is rated 2000 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{2000}{220} = 9.09\) A

Q536. An appliance is rated 2000 W at 110 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{2000}{110} = 18.18\) A

Q537. An appliance is rated 100 W at 220 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{100}{220} = 0.45\) A

Q538. An appliance is rated 500 W at 230 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{500}{230} = 2.17\) A

Q539. An appliance is rated 1500 W at 110 V. Find the current drawn.

Difficulty: Easy-Medium · Topic: Electrical Power

Solution

\(P = VI\), so \(I = \frac{P}{V}\)

\(I = \frac{1500}{110} = 13.64\) A

Q540. A 1000 W heater is used for 1 hours daily for 29 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 1\text{ h/day} \times 29\text{ days}\)

\(= 1000 \times 1 \times 29 = \) ... Wh

In kWh: divide by 1000 \(= 29.0\) kWh

Q541. A 2500 W heater is used for 8 hours daily for 27 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2500\text{ W} \times 8\text{ h/day} \times 27\text{ days}\)

\(= 2500 \times 8 \times 27 = \) ... Wh

In kWh: divide by 1000 \(= 540.0\) kWh

Q542. A 500 W heater is used for 2 hours daily for 28 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 2\text{ h/day} \times 28\text{ days}\)

\(= 500 \times 2 \times 28 = \) ... Wh

In kWh: divide by 1000 \(= 28.0\) kWh

Q543. A 1000 W heater is used for 1 hours daily for 9 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 1\text{ h/day} \times 9\text{ days}\)

\(= 1000 \times 1 \times 9 = \) ... Wh

In kWh: divide by 1000 \(= 9.0\) kWh

Q544. A 1000 W heater is used for 3 hours daily for 21 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 3\text{ h/day} \times 21\text{ days}\)

\(= 1000 \times 3 \times 21 = \) ... Wh

In kWh: divide by 1000 \(= 63.0\) kWh

Q545. A 3000 W heater is used for 2 hours daily for 14 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 2\text{ h/day} \times 14\text{ days}\)

\(= 3000 \times 2 \times 14 = \) ... Wh

In kWh: divide by 1000 \(= 84.0\) kWh

Q546. A 3000 W heater is used for 3 hours daily for 26 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 3\text{ h/day} \times 26\text{ days}\)

\(= 3000 \times 3 \times 26 = \) ... Wh

In kWh: divide by 1000 \(= 234.0\) kWh

Q547. A 2000 W heater is used for 4 hours daily for 23 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 4\text{ h/day} \times 23\text{ days}\)

\(= 2000 \times 4 \times 23 = \) ... Wh

In kWh: divide by 1000 \(= 184.0\) kWh

Q548. A 2500 W heater is used for 1 hours daily for 15 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2500\text{ W} \times 1\text{ h/day} \times 15\text{ days}\)

\(= 2500 \times 1 \times 15 = \) ... Wh

In kWh: divide by 1000 \(= 37.5\) kWh

Q549. A 1500 W heater is used for 6 hours daily for 26 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 6\text{ h/day} \times 26\text{ days}\)

\(= 1500 \times 6 \times 26 = \) ... Wh

In kWh: divide by 1000 \(= 234.0\) kWh

Q550. A 3000 W heater is used for 4 hours daily for 13 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 4\text{ h/day} \times 13\text{ days}\)

\(= 3000 \times 4 \times 13 = \) ... Wh

In kWh: divide by 1000 \(= 156.0\) kWh

Q551. A 1500 W heater is used for 7 hours daily for 24 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 7\text{ h/day} \times 24\text{ days}\)

\(= 1500 \times 7 \times 24 = \) ... Wh

In kWh: divide by 1000 \(= 252.0\) kWh

Q552. A 1500 W heater is used for 7 hours daily for 10 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 7\text{ h/day} \times 10\text{ days}\)

\(= 1500 \times 7 \times 10 = \) ... Wh

In kWh: divide by 1000 \(= 105.0\) kWh

Q553. A 2000 W heater is used for 7 hours daily for 21 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 7\text{ h/day} \times 21\text{ days}\)

\(= 2000 \times 7 \times 21 = \) ... Wh

In kWh: divide by 1000 \(= 294.0\) kWh

Q554. A 500 W heater is used for 5 hours daily for 13 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 5\text{ h/day} \times 13\text{ days}\)

\(= 500 \times 5 \times 13 = \) ... Wh

In kWh: divide by 1000 \(= 32.5\) kWh

Q555. A 1000 W heater is used for 1 hours daily for 25 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 1\text{ h/day} \times 25\text{ days}\)

\(= 1000 \times 1 \times 25 = \) ... Wh

In kWh: divide by 1000 \(= 25.0\) kWh

Q556. A 3000 W heater is used for 4 hours daily for 30 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 4\text{ h/day} \times 30\text{ days}\)

\(= 3000 \times 4 \times 30 = \) ... Wh

In kWh: divide by 1000 \(= 360.0\) kWh

Q557. A 2000 W heater is used for 6 hours daily for 24 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 6\text{ h/day} \times 24\text{ days}\)

\(= 2000 \times 6 \times 24 = \) ... Wh

In kWh: divide by 1000 \(= 288.0\) kWh

Q558. A 3000 W heater is used for 8 hours daily for 8 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 8\text{ h/day} \times 8\text{ days}\)

\(= 3000 \times 8 \times 8 = \) ... Wh

In kWh: divide by 1000 \(= 192.0\) kWh

Q559. A 3000 W heater is used for 1 hours daily for 7 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 1\text{ h/day} \times 7\text{ days}\)

\(= 3000 \times 1 \times 7 = \) ... Wh

In kWh: divide by 1000 \(= 21.0\) kWh

Q560. A 500 W heater is used for 4 hours daily for 17 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 4\text{ h/day} \times 17\text{ days}\)

\(= 500 \times 4 \times 17 = \) ... Wh

In kWh: divide by 1000 \(= 34.0\) kWh

Q561. A 500 W heater is used for 1 hours daily for 22 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 1\text{ h/day} \times 22\text{ days}\)

\(= 500 \times 1 \times 22 = \) ... Wh

In kWh: divide by 1000 \(= 11.0\) kWh

Q562. A 2000 W heater is used for 6 hours daily for 26 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 6\text{ h/day} \times 26\text{ days}\)

\(= 2000 \times 6 \times 26 = \) ... Wh

In kWh: divide by 1000 \(= 312.0\) kWh

Q563. A 3000 W heater is used for 6 hours daily for 5 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 6\text{ h/day} \times 5\text{ days}\)

\(= 3000 \times 6 \times 5 = \) ... Wh

In kWh: divide by 1000 \(= 90.0\) kWh

Q564. A 2000 W heater is used for 3 hours daily for 9 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 3\text{ h/day} \times 9\text{ days}\)

\(= 2000 \times 3 \times 9 = \) ... Wh

In kWh: divide by 1000 \(= 54.0\) kWh

Q565. A 2000 W heater is used for 8 hours daily for 15 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 8\text{ h/day} \times 15\text{ days}\)

\(= 2000 \times 8 \times 15 = \) ... Wh

In kWh: divide by 1000 \(= 240.0\) kWh

Q566. A 1500 W heater is used for 4 hours daily for 28 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 4\text{ h/day} \times 28\text{ days}\)

\(= 1500 \times 4 \times 28 = \) ... Wh

In kWh: divide by 1000 \(= 168.0\) kWh

Q567. A 2000 W heater is used for 8 hours daily for 21 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 8\text{ h/day} \times 21\text{ days}\)

\(= 2000 \times 8 \times 21 = \) ... Wh

In kWh: divide by 1000 \(= 336.0\) kWh

Q568. A 1000 W heater is used for 2 hours daily for 22 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 2\text{ h/day} \times 22\text{ days}\)

\(= 1000 \times 2 \times 22 = \) ... Wh

In kWh: divide by 1000 \(= 44.0\) kWh

Q569. A 1000 W heater is used for 5 hours daily for 11 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 5\text{ h/day} \times 11\text{ days}\)

\(= 1000 \times 5 \times 11 = \) ... Wh

In kWh: divide by 1000 \(= 55.0\) kWh

Q570. A 1500 W heater is used for 3 hours daily for 12 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 3\text{ h/day} \times 12\text{ days}\)

\(= 1500 \times 3 \times 12 = \) ... Wh

In kWh: divide by 1000 \(= 54.0\) kWh

Q571. A 3000 W heater is used for 7 hours daily for 11 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 7\text{ h/day} \times 11\text{ days}\)

\(= 3000 \times 7 \times 11 = \) ... Wh

In kWh: divide by 1000 \(= 231.0\) kWh

Q572. A 3000 W heater is used for 2 hours daily for 24 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 2\text{ h/day} \times 24\text{ days}\)

\(= 3000 \times 2 \times 24 = \) ... Wh

In kWh: divide by 1000 \(= 144.0\) kWh

Q573. A 2000 W heater is used for 8 hours daily for 24 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 8\text{ h/day} \times 24\text{ days}\)

\(= 2000 \times 8 \times 24 = \) ... Wh

In kWh: divide by 1000 \(= 384.0\) kWh

Q574. A 500 W heater is used for 7 hours daily for 10 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 7\text{ h/day} \times 10\text{ days}\)

\(= 500 \times 7 \times 10 = \) ... Wh

In kWh: divide by 1000 \(= 35.0\) kWh

Q575. A 3000 W heater is used for 5 hours daily for 8 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 5\text{ h/day} \times 8\text{ days}\)

\(= 3000 \times 5 \times 8 = \) ... Wh

In kWh: divide by 1000 \(= 120.0\) kWh

Q576. A 1000 W heater is used for 7 hours daily for 10 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 7\text{ h/day} \times 10\text{ days}\)

\(= 1000 \times 7 \times 10 = \) ... Wh

In kWh: divide by 1000 \(= 70.0\) kWh

Q577. A 1000 W heater is used for 1 hours daily for 22 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 1\text{ h/day} \times 22\text{ days}\)

\(= 1000 \times 1 \times 22 = \) ... Wh

In kWh: divide by 1000 \(= 22.0\) kWh

Q578. A 2000 W heater is used for 7 hours daily for 27 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 7\text{ h/day} \times 27\text{ days}\)

\(= 2000 \times 7 \times 27 = \) ... Wh

In kWh: divide by 1000 \(= 378.0\) kWh

Q579. A 3000 W heater is used for 1 hours daily for 28 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 1\text{ h/day} \times 28\text{ days}\)

\(= 3000 \times 1 \times 28 = \) ... Wh

In kWh: divide by 1000 \(= 84.0\) kWh

Q580. A 1500 W heater is used for 2 hours daily for 12 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 2\text{ h/day} \times 12\text{ days}\)

\(= 1500 \times 2 \times 12 = \) ... Wh

In kWh: divide by 1000 \(= 36.0\) kWh

Q581. A 1500 W heater is used for 6 hours daily for 14 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 6\text{ h/day} \times 14\text{ days}\)

\(= 1500 \times 6 \times 14 = \) ... Wh

In kWh: divide by 1000 \(= 126.0\) kWh

Q582. A 1500 W heater is used for 3 hours daily for 26 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 3\text{ h/day} \times 26\text{ days}\)

\(= 1500 \times 3 \times 26 = \) ... Wh

In kWh: divide by 1000 \(= 117.0\) kWh

Q583. A 3000 W heater is used for 4 hours daily for 18 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 4\text{ h/day} \times 18\text{ days}\)

\(= 3000 \times 4 \times 18 = \) ... Wh

In kWh: divide by 1000 \(= 216.0\) kWh

Q584. A 1000 W heater is used for 4 hours daily for 22 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 4\text{ h/day} \times 22\text{ days}\)

\(= 1000 \times 4 \times 22 = \) ... Wh

In kWh: divide by 1000 \(= 88.0\) kWh

Q585. A 3000 W heater is used for 2 hours daily for 29 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 2\text{ h/day} \times 29\text{ days}\)

\(= 3000 \times 2 \times 29 = \) ... Wh

In kWh: divide by 1000 \(= 174.0\) kWh

Q586. A 2000 W heater is used for 7 hours daily for 13 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 7\text{ h/day} \times 13\text{ days}\)

\(= 2000 \times 7 \times 13 = \) ... Wh

In kWh: divide by 1000 \(= 182.0\) kWh

Q587. A 2000 W heater is used for 4 hours daily for 27 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 4\text{ h/day} \times 27\text{ days}\)

\(= 2000 \times 4 \times 27 = \) ... Wh

In kWh: divide by 1000 \(= 216.0\) kWh

Q588. A 500 W heater is used for 5 hours daily for 29 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 5\text{ h/day} \times 29\text{ days}\)

\(= 500 \times 5 \times 29 = \) ... Wh

In kWh: divide by 1000 \(= 72.5\) kWh

Q589. A 1500 W heater is used for 6 hours daily for 13 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 6\text{ h/day} \times 13\text{ days}\)

\(= 1500 \times 6 \times 13 = \) ... Wh

In kWh: divide by 1000 \(= 117.0\) kWh

Q590. A 2000 W heater is used for 2 hours daily for 10 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 2\text{ h/day} \times 10\text{ days}\)

\(= 2000 \times 2 \times 10 = \) ... Wh

In kWh: divide by 1000 \(= 40.0\) kWh

Q591. A 2000 W heater is used for 6 hours daily for 7 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2000\text{ W} \times 6\text{ h/day} \times 7\text{ days}\)

\(= 2000 \times 6 \times 7 = \) ... Wh

In kWh: divide by 1000 \(= 84.0\) kWh

Q592. A 1000 W heater is used for 5 hours daily for 14 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1000\text{ W} \times 5\text{ h/day} \times 14\text{ days}\)

\(= 1000 \times 5 \times 14 = \) ... Wh

In kWh: divide by 1000 \(= 70.0\) kWh

Q593. A 500 W heater is used for 8 hours daily for 5 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 500\text{ W} \times 8\text{ h/day} \times 5\text{ days}\)

\(= 500 \times 8 \times 5 = \) ... Wh

In kWh: divide by 1000 \(= 20.0\) kWh

Q594. A 1500 W heater is used for 1 hours daily for 15 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 1\text{ h/day} \times 15\text{ days}\)

\(= 1500 \times 1 \times 15 = \) ... Wh

In kWh: divide by 1000 \(= 22.5\) kWh

Q595. A 1500 W heater is used for 2 hours daily for 19 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 1500\text{ W} \times 2\text{ h/day} \times 19\text{ days}\)

\(= 1500 \times 2 \times 19 = \) ... Wh

In kWh: divide by 1000 \(= 57.0\) kWh

Q596. A 3000 W heater is used for 4 hours daily for 14 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 3000\text{ W} \times 4\text{ h/day} \times 14\text{ days}\)

\(= 3000 \times 4 \times 14 = \) ... Wh

In kWh: divide by 1000 \(= 168.0\) kWh

Q597. A 2500 W heater is used for 7 hours daily for 24 days. Find the total energy consumed in kWh.

Difficulty: Easy-Medium · Topic: Electrical Energy (kWh)

Solution

Energy = Power x Time

\(= 2500\text{ W} \times 7\text{ h/day} \times 24\text{ days}\)

\(= 2500 \times 7 \times 24 = \) ... Wh

In kWh: divide by 1000 \(= 420.0\) kWh

Q598. A current of 3.0 A flows through a resistance of 15 ohm for 83 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 15 \times 83 = 11205.0\) J

Q599. A current of 5.0 A flows through a resistance of 5 ohm for 74 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 5 \times 74 = 9250.0\) J

Q600. A current of 4.0 A flows through a resistance of 32 ohm for 55 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 32 \times 55 = 28160.0\) J

Q601. A current of 1.0 A flows through a resistance of 43 ohm for 84 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 43 \times 84 = 3612.0\) J

Q602. A current of 2.0 A flows through a resistance of 47 ohm for 54 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 47 \times 54 = 10152.0\) J

Q603. A current of 1.0 A flows through a resistance of 38 ohm for 20 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 38 \times 20 = 760.0\) J

Q604. A current of 1.0 A flows through a resistance of 39 ohm for 100 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 39 \times 100 = 3900.0\) J

Q605. A current of 5.0 A flows through a resistance of 47 ohm for 20 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 47 \times 20 = 23500.0\) J

Q606. A current of 4.0 A flows through a resistance of 23 ohm for 46 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 23 \times 46 = 16928.0\) J

Q607. A current of 0.5 A flows through a resistance of 24 ohm for 72 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 24 \times 72 = 432.0\) J

Q608. A current of 0.5 A flows through a resistance of 20 ohm for 27 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 20 \times 27 = 135.0\) J

Q609. A current of 2.0 A flows through a resistance of 30 ohm for 94 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 30 \times 94 = 11280.0\) J

Q610. A current of 3.0 A flows through a resistance of 8 ohm for 74 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 8 \times 74 = 5328.0\) J

Q611. A current of 2.0 A flows through a resistance of 46 ohm for 108 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 46 \times 108 = 19872.0\) J

Q612. A current of 2.0 A flows through a resistance of 19 ohm for 72 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 19 \times 72 = 5472.0\) J

Q613. A current of 3.0 A flows through a resistance of 8 ohm for 21 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 8 \times 21 = 1512.0\) J

Q614. A current of 0.5 A flows through a resistance of 37 ohm for 64 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 37 \times 64 = 592.0\) J

Q615. A current of 3.0 A flows through a resistance of 50 ohm for 95 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 50 \times 95 = 42750.0\) J

Q616. A current of 2.0 A flows through a resistance of 15 ohm for 112 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 15 \times 112 = 6720.0\) J

Q617. A current of 4.0 A flows through a resistance of 24 ohm for 81 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 24 \times 81 = 31104.0\) J

Q618. A current of 4.0 A flows through a resistance of 40 ohm for 52 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 40 \times 52 = 33280.0\) J

Q619. A current of 1.0 A flows through a resistance of 49 ohm for 104 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 49 \times 104 = 5096.0\) J

Q620. A current of 2.0 A flows through a resistance of 46 ohm for 18 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 46 \times 18 = 3312.0\) J

Q621. A current of 3.0 A flows through a resistance of 37 ohm for 69 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 37 \times 69 = 22977.0\) J

Q622. A current of 2.0 A flows through a resistance of 29 ohm for 99 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 29 \times 99 = 11484.0\) J

Q623. A current of 5.0 A flows through a resistance of 44 ohm for 57 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 44 \times 57 = 62700.0\) J

Q624. A current of 1.0 A flows through a resistance of 22 ohm for 76 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 22 \times 76 = 1672.0\) J

Q625. A current of 3.0 A flows through a resistance of 26 ohm for 88 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 26 \times 88 = 20592.0\) J

Q626. A current of 0.5 A flows through a resistance of 42 ohm for 113 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 42 \times 113 = 1186.5\) J

Q627. A current of 4.0 A flows through a resistance of 34 ohm for 43 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 34 \times 43 = 23392.0\) J

Q628. A current of 1.0 A flows through a resistance of 14 ohm for 108 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 14 \times 108 = 1512.0\) J

Q629. A current of 0.5 A flows through a resistance of 6 ohm for 91 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 6 \times 91 = 136.5\) J

Q630. A current of 3.0 A flows through a resistance of 7 ohm for 42 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 7 \times 42 = 2646.0\) J

Q631. A current of 2.0 A flows through a resistance of 30 ohm for 67 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 30 \times 67 = 8040.0\) J

Q632. A current of 5.0 A flows through a resistance of 24 ohm for 100 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 24 \times 100 = 60000.0\) J

Q633. A current of 3.0 A flows through a resistance of 46 ohm for 116 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 46 \times 116 = 48024.0\) J

Q634. A current of 5.0 A flows through a resistance of 33 ohm for 47 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 33 \times 47 = 38775.0\) J

Q635. A current of 1.0 A flows through a resistance of 7 ohm for 83 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 7 \times 83 = 581.0\) J

Q636. A current of 2.0 A flows through a resistance of 9 ohm for 39 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 9 \times 39 = 1404.0\) J

Q637. A current of 5.0 A flows through a resistance of 25 ohm for 65 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 25 \times 65 = 40625.0\) J

Q638. A current of 5.0 A flows through a resistance of 9 ohm for 70 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 9 \times 70 = 15750.0\) J

Q639. A current of 2.0 A flows through a resistance of 27 ohm for 92 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 27 \times 92 = 9936.0\) J

Q640. A current of 5.0 A flows through a resistance of 9 ohm for 115 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 9 \times 115 = 25875.0\) J

Q641. A current of 5.0 A flows through a resistance of 14 ohm for 118 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 14 \times 118 = 41300.0\) J

Q642. A current of 1.0 A flows through a resistance of 44 ohm for 14 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 44 \times 14 = 616.0\) J

Q643. A current of 2.0 A flows through a resistance of 17 ohm for 57 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 17 \times 57 = 3876.0\) J

Q644. A current of 2.0 A flows through a resistance of 32 ohm for 69 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 32 \times 69 = 8832.0\) J

Q645. A current of 4.0 A flows through a resistance of 34 ohm for 85 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 34 \times 85 = 46240.0\) J

Q646. A current of 5.0 A flows through a resistance of 40 ohm for 50 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (5.0)^2 \times 40 \times 50 = 50000.0\) J

Q647. A current of 0.5 A flows through a resistance of 14 ohm for 96 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 14 \times 96 = 336.0\) J

Q648. A current of 0.5 A flows through a resistance of 48 ohm for 52 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 48 \times 52 = 624.0\) J

Q649. A current of 2.0 A flows through a resistance of 32 ohm for 92 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 32 \times 92 = 11776.0\) J

Q650. A current of 1.0 A flows through a resistance of 7 ohm for 52 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (1.0)^2 \times 7 \times 52 = 364.0\) J

Q651. A current of 3.0 A flows through a resistance of 21 ohm for 54 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (3.0)^2 \times 21 \times 54 = 10206.0\) J

Q652. A current of 0.5 A flows through a resistance of 26 ohm for 97 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 26 \times 97 = 630.5\) J

Q653. A current of 2.0 A flows through a resistance of 43 ohm for 74 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 43 \times 74 = 12728.0\) J

Q654. A current of 2.0 A flows through a resistance of 9 ohm for 60 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (2.0)^2 \times 9 \times 60 = 2160.0\) J

Q655. A current of 0.5 A flows through a resistance of 27 ohm for 70 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 27 \times 70 = 472.5\) J

Q656. A current of 0.5 A flows through a resistance of 49 ohm for 52 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (0.5)^2 \times 49 \times 52 = 637.0\) J

Q657. A current of 4.0 A flows through a resistance of 30 ohm for 73 s. Find the heat produced.

Difficulty: Easy-Medium · Topic: Joule's Law of Heating (H=I^2Rt)

Solution

\(H = I^2 R t = (4.0)^2 \times 30 \times 73 = 35040.0\) J

Q658. A wire of resistance 28 ohm has length 0.5 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{28 \times 2.0 \times 10^{-6}}{0.5} = 1.12e-04\) ohm.m

Q659. A wire of resistance 18 ohm has length 5 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{18 \times 0.1 \times 10^{-6}}{5} = 3.60e-07\) ohm.m

Q660. A wire of resistance 42 ohm has length 5 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{42 \times 2.0 \times 10^{-6}}{5} = 1.68e-05\) ohm.m

Q661. A wire of resistance 37 ohm has length 0.5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{37 \times 1.5 \times 10^{-6}}{0.5} = 1.11e-04\) ohm.m

Q662. A wire of resistance 29 ohm has length 0.5 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{29 \times 0.1 \times 10^{-6}}{0.5} = 5.80e-06\) ohm.m

Q663. A wire of resistance 10 ohm has length 1 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{10 \times 1.5 \times 10^{-6}}{1} = 1.50e-05\) ohm.m

Q664. A wire of resistance 44 ohm has length 5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{44 \times 1.5 \times 10^{-6}}{5} = 1.32e-05\) ohm.m

Q665. A wire of resistance 23 ohm has length 1 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{23 \times 0.5 \times 10^{-6}}{1} = 1.15e-05\) ohm.m

Q666. A wire of resistance 11 ohm has length 3 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{11 \times 2.0 \times 10^{-6}}{3} = 7.33e-06\) ohm.m

Q667. A wire of resistance 7 ohm has length 10 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{7 \times 0.5 \times 10^{-6}}{10} = 3.50e-07\) ohm.m

Q668. A wire of resistance 48 ohm has length 3 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{48 \times 0.25 \times 10^{-6}}{3} = 4.00e-06\) ohm.m

Q669. A wire of resistance 21 ohm has length 1 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{21 \times 1.0 \times 10^{-6}}{1} = 2.10e-05\) ohm.m

Q670. A wire of resistance 28 ohm has length 0.5 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{28 \times 1.0 \times 10^{-6}}{0.5} = 5.60e-05\) ohm.m

Q671. A wire of resistance 33 ohm has length 5 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{33 \times 0.5 \times 10^{-6}}{5} = 3.30e-06\) ohm.m

Q672. A wire of resistance 5 ohm has length 1 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{5 \times 1.0 \times 10^{-6}}{1} = 5.00e-06\) ohm.m

Q673. A wire of resistance 9 ohm has length 3 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{9 \times 0.1 \times 10^{-6}}{3} = 3.00e-07\) ohm.m

Q674. A wire of resistance 10 ohm has length 1 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{10 \times 2.0 \times 10^{-6}}{1} = 2.00e-05\) ohm.m

Q675. A wire of resistance 31 ohm has length 1 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{31 \times 0.25 \times 10^{-6}}{1} = 7.75e-06\) ohm.m

Q676. A wire of resistance 25 ohm has length 5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{25 \times 1.5 \times 10^{-6}}{5} = 7.50e-06\) ohm.m

Q677. A wire of resistance 21 ohm has length 2 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{21 \times 0.5 \times 10^{-6}}{2} = 5.25e-06\) ohm.m

Q678. A wire of resistance 18 ohm has length 5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{18 \times 1.5 \times 10^{-6}}{5} = 5.40e-06\) ohm.m

Q679. A wire of resistance 15 ohm has length 2 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{15 \times 1.5 \times 10^{-6}}{2} = 1.12e-05\) ohm.m

Q680. A wire of resistance 2 ohm has length 1 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{2 \times 1.5 \times 10^{-6}}{1} = 3.00e-06\) ohm.m

Q681. A wire of resistance 42 ohm has length 10 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{42 \times 0.25 \times 10^{-6}}{10} = 1.05e-06\) ohm.m

Q682. A wire of resistance 30 ohm has length 2 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{30 \times 2.0 \times 10^{-6}}{2} = 3.00e-05\) ohm.m

Q683. A wire of resistance 40 ohm has length 2 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{40 \times 1.5 \times 10^{-6}}{2} = 3.00e-05\) ohm.m

Q684. A wire of resistance 7 ohm has length 5 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{7 \times 0.1 \times 10^{-6}}{5} = 1.40e-07\) ohm.m

Q685. A wire of resistance 25 ohm has length 10 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{25 \times 1.5 \times 10^{-6}}{10} = 3.75e-06\) ohm.m

Q686. A wire of resistance 28 ohm has length 5 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{28 \times 2.0 \times 10^{-6}}{5} = 1.12e-05\) ohm.m

Q687. A wire of resistance 9 ohm has length 2 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{9 \times 2.0 \times 10^{-6}}{2} = 9.00e-06\) ohm.m

Q688. A wire of resistance 3 ohm has length 1 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{3 \times 0.25 \times 10^{-6}}{1} = 7.50e-07\) ohm.m

Q689. A wire of resistance 7 ohm has length 1 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{7 \times 0.1 \times 10^{-6}}{1} = 7.00e-07\) ohm.m

Q690. A wire of resistance 37 ohm has length 3 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{37 \times 0.25 \times 10^{-6}}{3} = 3.08e-06\) ohm.m

Q691. A wire of resistance 28 ohm has length 1 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{28 \times 0.5 \times 10^{-6}}{1} = 1.40e-05\) ohm.m

Q692. A wire of resistance 13 ohm has length 10 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{13 \times 1.0 \times 10^{-6}}{10} = 1.30e-06\) ohm.m

Q693. A wire of resistance 34 ohm has length 3 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{34 \times 0.1 \times 10^{-6}}{3} = 1.13e-06\) ohm.m

Q694. A wire of resistance 9 ohm has length 3 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{9 \times 0.25 \times 10^{-6}}{3} = 7.50e-07\) ohm.m

Q695. A wire of resistance 43 ohm has length 5 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{43 \times 0.1 \times 10^{-6}}{5} = 8.60e-07\) ohm.m

Q696. A wire of resistance 5 ohm has length 0.5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{5 \times 1.5 \times 10^{-6}}{0.5} = 1.50e-05\) ohm.m

Q697. A wire of resistance 49 ohm has length 3 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{49 \times 1.0 \times 10^{-6}}{3} = 1.63e-05\) ohm.m

Q698. A wire of resistance 42 ohm has length 1 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{42 \times 0.25 \times 10^{-6}}{1} = 1.05e-05\) ohm.m

Q699. A wire of resistance 31 ohm has length 5 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{31 \times 2.0 \times 10^{-6}}{5} = 1.24e-05\) ohm.m

Q700. A wire of resistance 41 ohm has length 1 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{41 \times 0.5 \times 10^{-6}}{1} = 2.05e-05\) ohm.m

Q701. A wire of resistance 14 ohm has length 0.5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{14 \times 1.5 \times 10^{-6}}{0.5} = 4.20e-05\) ohm.m

Q702. A wire of resistance 45 ohm has length 1 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{45 \times 2.0 \times 10^{-6}}{1} = 9.00e-05\) ohm.m

Q703. A wire of resistance 27 ohm has length 10 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{27 \times 0.1 \times 10^{-6}}{10} = 2.70e-07\) ohm.m

Q704. A wire of resistance 33 ohm has length 1 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{33 \times 1.0 \times 10^{-6}}{1} = 3.30e-05\) ohm.m

Q705. A wire of resistance 48 ohm has length 0.5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{48 \times 1.5 \times 10^{-6}}{0.5} = 1.44e-04\) ohm.m

Q706. A wire of resistance 33 ohm has length 2 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{33 \times 0.25 \times 10^{-6}}{2} = 4.12e-06\) ohm.m

Q707. A wire of resistance 39 ohm has length 1 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{39 \times 1.0 \times 10^{-6}}{1} = 3.90e-05\) ohm.m

Q708. A wire of resistance 40 ohm has length 1 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{40 \times 0.1 \times 10^{-6}}{1} = 4.00e-06\) ohm.m

Q709. A wire of resistance 13 ohm has length 0.5 m and cross-sectional area 1.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.0 mm^2 = 1.0 x 10^-6 m^2

\(\rho = \frac{13 \times 1.0 \times 10^{-6}}{0.5} = 2.60e-05\) ohm.m

Q710. A wire of resistance 22 ohm has length 5 m and cross-sectional area 0.1 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.1 mm^2 = 0.1 x 10^-6 m^2

\(\rho = \frac{22 \times 0.1 \times 10^{-6}}{5} = 4.40e-07\) ohm.m

Q711. A wire of resistance 35 ohm has length 10 m and cross-sectional area 0.25 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.25 mm^2 = 0.25 x 10^-6 m^2

\(\rho = \frac{35 \times 0.25 \times 10^{-6}}{10} = 8.75e-07\) ohm.m

Q712. A wire of resistance 20 ohm has length 3 m and cross-sectional area 0.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 0.5 mm^2 = 0.5 x 10^-6 m^2

\(\rho = \frac{20 \times 0.5 \times 10^{-6}}{3} = 3.33e-06\) ohm.m

Q713. A wire of resistance 11 ohm has length 3 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{11 \times 1.5 \times 10^{-6}}{3} = 5.50e-06\) ohm.m

Q714. A wire of resistance 34 ohm has length 3 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{34 \times 2.0 \times 10^{-6}}{3} = 2.27e-05\) ohm.m

Q715. A wire of resistance 27 ohm has length 0.5 m and cross-sectional area 2.0 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 2.0 mm^2 = 2.0 x 10^-6 m^2

\(\rho = \frac{27 \times 2.0 \times 10^{-6}}{0.5} = 1.08e-04\) ohm.m

Q716. A wire of resistance 9 ohm has length 5 m and cross-sectional area 1.5 mm^2. Find the resistivity.

Difficulty: Easy-Medium · Topic: Resistivity of a material

Solution

\(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l}\)

A = 1.5 mm^2 = 1.5 x 10^-6 m^2

\(\rho = \frac{9 \times 1.5 \times 10^{-6}}{5} = 2.70e-06\) ohm.m

Q717. How many electrons constitute a current of 1 A flowing for 1 second? (Charge on electron = 1.6 × 10⁻¹⁹ C)

Difficulty: Medium · Topic: Electric Current

Solution

Given: I = 1 A, t = 1 s, e = 1.6 × 10⁻¹⁹ C

Step 1: Total charge Q = I × t = 1 × 1 = 1 C

Step 2: Number of electrons n = Q/e

n = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons

So about 6.25 billion billion electrons flow past a point every second when the current is 1 A!

Q718. A wire of length 2 m and cross-sectional area 1 × 10⁻⁶ m² has a resistance of 10 Ω. Calculate the resistivity of the material of the wire.

Difficulty: Medium · Topic: Resistance

Solution

Given: l = 2 m, A = 1 × 10⁻⁶ m², R = 10 Ω

Formula: R = ρl/A → ρ = RA/l

ρ = (10 × 1 × 10⁻⁶) / 2

ρ = 10 × 10⁻⁶ / 2

ρ = 5 × 10⁻⁶ Ω·m

Q719. A wire of resistance R is stretched to double its length. Its new resistance is:

Difficulty: Medium · Topic: Resistance

Solution

When a wire is stretched, its volume remains constant.

Original: R = ρl/A, Volume = l × A

After stretching: new length l' = 2l. Since volume is constant:

l × A = l' × A' → A × l = 2l × A' → A' = A/2

New resistance: R' = ρl'/A' = ρ(2l)/(A/2) = 4ρl/A = 4R

The resistance becomes 4 times. In general, if length is made n times, resistance becomes n² times.

Q720. Three resistors of 2 Ω, 3 Ω, and 5 Ω are connected in series to a 20 V battery. Find: (a) the total resistance, (b) the current through the circuit, and (c) the voltage drop across each resistor.

Difficulty: Medium · Topic: Resistors in Series

Solution

(a) Total resistance:

R = R₁ + R₂ + R₃ = 2 + 3 + 5 = 10 Ω

(b) Current (same throughout in series):

I = V/R = 20/10 = 2 A

(c) Voltage drops:

V₁ = IR₁ = 2 × 2 = 4 V

V₂ = IR₂ = 2 × 3 = 6 V

V₃ = IR₃ = 2 × 5 = 10 V

Verification: V₁ + V₂ + V₃ = 4 + 6 + 10 = 20 V ✓ (equals the battery voltage)

Q721. Two resistors of 6 Ω and 12 Ω are connected in parallel. Find the equivalent resistance and the total current if a 12 V battery is connected across them.

Difficulty: Medium · Topic: Resistors in Parallel

Solution

Equivalent resistance:

1/R_eq = 1/R₁ + 1/R₂ = 1/6 + 1/12 = (2 + 1)/12 = 3/12 = 1/4

R_eq = 4 Ω

Note: 4 Ω is less than the smallest resistor (6 Ω), as expected for parallel combinations.

Total current:

I = V/R_eq = 12/4 = 3 A

Individual currents:

I₁ = V/R₁ = 12/6 = 2 A

I₂ = V/R₂ = 12/12 = 1 A

Verification: I₁ + I₂ = 2 + 1 = 3 A ✓ (equals total current)

Q722. An electric iron consumes energy at the rate of 1000 W. If the current drawn is 5 A, what is the resistance of its heating element? Also, calculate the heat produced in 30 seconds.

Difficulty: Medium · Topic: Heating Effect

Solution

Given: P = 1000 W, I = 5 A

Step 1: Find resistance

P = I²R → R = P/I² = 1000/(5)² = 1000/25 = 40 Ω

Step 2: Heat produced in 30 seconds

H = I²Rt = (5)² × 40 × 30 = 25 × 40 × 30 = 30,000 J = 30 kJ

Alternatively: H = P × t = 1000 × 30 = 30,000 J ✓

Q723. An electric heater rated 1500 W is used for 2 hours daily. Calculate the cost of using it for 30 days if the cost of 1 unit (kWh) of electricity is ₹5.

Difficulty: Medium · Topic: Electric Power

Solution

Given: P = 1500 W = 1.5 kW, t = 2 hours/day, days = 30

Step 1: Energy consumed per day

E = P × t = 1.5 × 2 = 3 kWh per day

Step 2: Total energy in 30 days

E_total = 3 × 30 = 90 kWh = 90 units

Step 3: Cost

Cost = 90 × 5 = ₹450

Q724. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Difficulty: Medium · Topic: Heating Effect

Solution

Electric toasters and irons use alloy coils (typically nichrome) instead of pure metals for three reasons:

  • High resistivity: Alloys have much higher resistivity than pure metals. Higher resistance means more heat is produced (H = I²Rt). Pure metals like copper have very low resistance and would produce very little heat.
  • High melting point: The coils get very hot during operation. Alloys like nichrome have high melting points (~1400°C) and can withstand these temperatures without melting or breaking.
  • Stable resistance: The resistance of alloys does not change significantly with temperature, unlike pure metals. This ensures the device performs consistently.
Q725. A copper wire has a resistance of 0.5 Ω. Another copper wire of the same mass but double the length is taken. What is its resistance?

Difficulty: Medium · Topic: Resistance

Solution

Given: R₁ = 0.5 Ω, same mass, l₂ = 2l₁

Same mass and same material means same volume (since density is the same).

Volume = l × A = constant

l₁ × A₁ = l₂ × A₂ = 2l₁ × A₂

A₂ = A₁/2

New resistance:

R₂ = ρl₂/A₂ = ρ(2l₁)/(A₁/2) = 4ρl₁/A₁ = 4R₁ = 4 × 0.5 = 2 Ω

Q726. Why are household appliances connected in parallel rather than in series? Give three reasons.

Difficulty: Medium · Topic: Resistors in Parallel

Solution

Household appliances are connected in parallel for the following reasons:

  • Same voltage across each appliance: In a parallel circuit, each appliance gets the full supply voltage (e.g., 220 V). In series, the voltage would be divided among appliances, and each would get less than the supply voltage — they wouldn't work properly.
  • Independent operation: In a parallel circuit, each appliance can be switched on or off independently. If one appliance fails, the others continue to work. In series, if one appliance fails, all others stop working.
  • Lower equivalent resistance: Parallel connections have lower total resistance, allowing the circuit to draw adequate current. In series, adding more appliances increases resistance and reduces current to each, making them dimmer or slower.
Q727. A cell of EMF 2 V and internal resistance 0.5 ohm is connected to an external resistance of 27 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{27 + 0.5} = 0.07\) A

Q728. A cell of EMF 8 V and internal resistance 0.5 ohm is connected to an external resistance of 17 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{8}{17 + 0.5} = 0.46\) A

Q729. A cell of EMF 2 V and internal resistance 1.5 ohm is connected to an external resistance of 17 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{17 + 1.5} = 0.11\) A

Q730. A cell of EMF 9 V and internal resistance 2.0 ohm is connected to an external resistance of 25 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{9}{25 + 2.0} = 0.33\) A

Q731. A cell of EMF 5 V and internal resistance 0.2 ohm is connected to an external resistance of 3 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{5}{3 + 0.2} = 1.56\) A

Q732. A cell of EMF 6 V and internal resistance 0.5 ohm is connected to an external resistance of 23 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{6}{23 + 0.5} = 0.26\) A

Q733. A cell of EMF 4 V and internal resistance 0.2 ohm is connected to an external resistance of 12 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{4}{12 + 0.2} = 0.33\) A

Q734. A cell of EMF 11 V and internal resistance 0.2 ohm is connected to an external resistance of 12 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{11}{12 + 0.2} = 0.9\) A

Q735. A cell of EMF 10 V and internal resistance 1.5 ohm is connected to an external resistance of 18 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{10}{18 + 1.5} = 0.51\) A

Q736. A cell of EMF 10 V and internal resistance 2.0 ohm is connected to an external resistance of 3 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{10}{3 + 2.0} = 2.0\) A

Q737. A cell of EMF 4 V and internal resistance 0.5 ohm is connected to an external resistance of 4 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{4}{4 + 0.5} = 0.89\) A

Q738. A cell of EMF 3 V and internal resistance 1.5 ohm is connected to an external resistance of 19 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{19 + 1.5} = 0.15\) A

Q739. A cell of EMF 1 V and internal resistance 2.0 ohm is connected to an external resistance of 18 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{1}{18 + 2.0} = 0.05\) A

Q740. A cell of EMF 8 V and internal resistance 0.5 ohm is connected to an external resistance of 24 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{8}{24 + 0.5} = 0.33\) A

Q741. A cell of EMF 9 V and internal resistance 1.5 ohm is connected to an external resistance of 18 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{9}{18 + 1.5} = 0.46\) A

Q742. A cell of EMF 3 V and internal resistance 0.5 ohm is connected to an external resistance of 11 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{11 + 0.5} = 0.26\) A

Q743. A cell of EMF 5 V and internal resistance 0.2 ohm is connected to an external resistance of 22 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{5}{22 + 0.2} = 0.23\) A

Q744. A cell of EMF 5 V and internal resistance 2.0 ohm is connected to an external resistance of 9 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{5}{9 + 2.0} = 0.45\) A

Q745. A cell of EMF 9 V and internal resistance 1.0 ohm is connected to an external resistance of 28 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{9}{28 + 1.0} = 0.31\) A

Q746. A cell of EMF 12 V and internal resistance 2.0 ohm is connected to an external resistance of 17 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{12}{17 + 2.0} = 0.63\) A

Q747. A cell of EMF 12 V and internal resistance 2.0 ohm is connected to an external resistance of 11 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{12}{11 + 2.0} = 0.92\) A

Q748. A cell of EMF 12 V and internal resistance 1.5 ohm is connected to an external resistance of 7 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{12}{7 + 1.5} = 1.41\) A

Q749. A cell of EMF 6 V and internal resistance 1.5 ohm is connected to an external resistance of 6 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{6}{6 + 1.5} = 0.8\) A

Q750. A cell of EMF 4 V and internal resistance 0.2 ohm is connected to an external resistance of 17 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{4}{17 + 0.2} = 0.23\) A

Q751. A cell of EMF 3 V and internal resistance 0.3 ohm is connected to an external resistance of 7 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{7 + 0.3} = 0.41\) A

Q752. A cell of EMF 2 V and internal resistance 2.0 ohm is connected to an external resistance of 9 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{9 + 2.0} = 0.18\) A

Q753. A cell of EMF 3 V and internal resistance 0.3 ohm is connected to an external resistance of 9 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{9 + 0.3} = 0.32\) A

Q754. A cell of EMF 8 V and internal resistance 0.2 ohm is connected to an external resistance of 14 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{8}{14 + 0.2} = 0.56\) A

Q755. A cell of EMF 3 V and internal resistance 0.2 ohm is connected to an external resistance of 11 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{11 + 0.2} = 0.27\) A

Q756. A cell of EMF 10 V and internal resistance 2.0 ohm is connected to an external resistance of 18 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{10}{18 + 2.0} = 0.5\) A

Q757. A cell of EMF 7 V and internal resistance 1.0 ohm is connected to an external resistance of 27 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{7}{27 + 1.0} = 0.25\) A

Q758. A cell of EMF 6 V and internal resistance 1.5 ohm is connected to an external resistance of 10 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{6}{10 + 1.5} = 0.52\) A

Q759. A cell of EMF 1 V and internal resistance 1.0 ohm is connected to an external resistance of 26 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{1}{26 + 1.0} = 0.04\) A

Q760. A cell of EMF 1 V and internal resistance 1.0 ohm is connected to an external resistance of 23 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{1}{23 + 1.0} = 0.04\) A

Q761. A cell of EMF 10 V and internal resistance 0.5 ohm is connected to an external resistance of 4 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{10}{4 + 0.5} = 2.22\) A

Q762. A cell of EMF 2 V and internal resistance 1.0 ohm is connected to an external resistance of 10 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{10 + 1.0} = 0.18\) A

Q763. A cell of EMF 3 V and internal resistance 1.0 ohm is connected to an external resistance of 7 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{7 + 1.0} = 0.38\) A

Q764. A cell of EMF 10 V and internal resistance 1.5 ohm is connected to an external resistance of 6 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{10}{6 + 1.5} = 1.33\) A

Q765. A cell of EMF 4 V and internal resistance 0.5 ohm is connected to an external resistance of 7 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{4}{7 + 0.5} = 0.53\) A

Q766. A cell of EMF 5 V and internal resistance 1.0 ohm is connected to an external resistance of 16 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{5}{16 + 1.0} = 0.29\) A

Q767. A cell of EMF 7 V and internal resistance 0.2 ohm is connected to an external resistance of 9 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{7}{9 + 0.2} = 0.76\) A

Q768. A cell of EMF 7 V and internal resistance 2.0 ohm is connected to an external resistance of 25 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{7}{25 + 2.0} = 0.26\) A

Q769. A cell of EMF 5 V and internal resistance 0.5 ohm is connected to an external resistance of 8 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{5}{8 + 0.5} = 0.59\) A

Q770. A cell of EMF 6 V and internal resistance 2.0 ohm is connected to an external resistance of 26 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{6}{26 + 2.0} = 0.21\) A

Q771. A cell of EMF 10 V and internal resistance 1.5 ohm is connected to an external resistance of 9 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{10}{9 + 1.5} = 0.95\) A

Q772. A cell of EMF 2 V and internal resistance 0.5 ohm is connected to an external resistance of 14 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{14 + 0.5} = 0.14\) A

Q773. A cell of EMF 7 V and internal resistance 1.0 ohm is connected to an external resistance of 20 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{7}{20 + 1.0} = 0.33\) A

Q774. A cell of EMF 7 V and internal resistance 0.3 ohm is connected to an external resistance of 24 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{7}{24 + 0.3} = 0.29\) A

Q775. A cell of EMF 1 V and internal resistance 1.5 ohm is connected to an external resistance of 21 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{1}{21 + 1.5} = 0.04\) A

Q776. A cell of EMF 2 V and internal resistance 1.5 ohm is connected to an external resistance of 15 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{15 + 1.5} = 0.12\) A

Q777. A cell of EMF 7 V and internal resistance 1.5 ohm is connected to an external resistance of 10 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{7}{10 + 1.5} = 0.61\) A

Q778. A cell of EMF 4 V and internal resistance 1.0 ohm is connected to an external resistance of 10 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{4}{10 + 1.0} = 0.36\) A

Q779. A cell of EMF 11 V and internal resistance 0.5 ohm is connected to an external resistance of 5 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{11}{5 + 0.5} = 2.0\) A

Q780. A cell of EMF 3 V and internal resistance 1.0 ohm is connected to an external resistance of 19 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{19 + 1.0} = 0.15\) A

Q781. A cell of EMF 8 V and internal resistance 1.5 ohm is connected to an external resistance of 8 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{8}{8 + 1.5} = 0.84\) A

Q782. A cell of EMF 10 V and internal resistance 1.5 ohm is connected to an external resistance of 16 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{10}{16 + 1.5} = 0.57\) A

Q783. A cell of EMF 2 V and internal resistance 2.0 ohm is connected to an external resistance of 28 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{28 + 2.0} = 0.07\) A

Q784. A cell of EMF 5 V and internal resistance 0.3 ohm is connected to an external resistance of 9 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{5}{9 + 0.3} = 0.54\) A

Q785. A cell of EMF 8 V and internal resistance 0.5 ohm is connected to an external resistance of 3 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{8}{3 + 0.5} = 2.29\) A

Q786. A cell of EMF 4 V and internal resistance 0.3 ohm is connected to an external resistance of 26 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{4}{26 + 0.3} = 0.15\) A

Q787. A cell of EMF 9 V and internal resistance 1.0 ohm is connected to an external resistance of 18 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{9}{18 + 1.0} = 0.47\) A

Q788. A cell of EMF 4 V and internal resistance 2.0 ohm is connected to an external resistance of 4 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{4}{4 + 2.0} = 0.67\) A

Q789. A cell of EMF 11 V and internal resistance 0.2 ohm is connected to an external resistance of 15 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{11}{15 + 0.2} = 0.72\) A

Q790. A cell of EMF 4 V and internal resistance 1.5 ohm is connected to an external resistance of 25 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{4}{25 + 1.5} = 0.15\) A

Q791. A cell of EMF 1 V and internal resistance 2.0 ohm is connected to an external resistance of 13 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{1}{13 + 2.0} = 0.07\) A

Q792. A cell of EMF 11 V and internal resistance 1.0 ohm is connected to an external resistance of 8 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{11}{8 + 1.0} = 1.22\) A

Q793. A cell of EMF 12 V and internal resistance 0.3 ohm is connected to an external resistance of 15 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{12}{15 + 0.3} = 0.78\) A

Q794. A cell of EMF 8 V and internal resistance 0.2 ohm is connected to an external resistance of 22 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{8}{22 + 0.2} = 0.36\) A

Q795. A cell of EMF 12 V and internal resistance 0.2 ohm is connected to an external resistance of 12 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{12}{12 + 0.2} = 0.98\) A

Q796. A cell of EMF 11 V and internal resistance 1.5 ohm is connected to an external resistance of 6 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{11}{6 + 1.5} = 1.47\) A

Q797. A cell of EMF 11 V and internal resistance 2.0 ohm is connected to an external resistance of 6 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{11}{6 + 2.0} = 1.38\) A

Q798. A cell of EMF 11 V and internal resistance 2.0 ohm is connected to an external resistance of 29 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{11}{29 + 2.0} = 0.35\) A

Q799. A cell of EMF 2 V and internal resistance 0.3 ohm is connected to an external resistance of 3 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{3 + 0.3} = 0.61\) A

Q800. A cell of EMF 8 V and internal resistance 2.0 ohm is connected to an external resistance of 28 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{8}{28 + 2.0} = 0.27\) A

Q801. A cell of EMF 3 V and internal resistance 2.0 ohm is connected to an external resistance of 18 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{18 + 2.0} = 0.15\) A

Q802. A cell of EMF 5 V and internal resistance 0.5 ohm is connected to an external resistance of 20 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{5}{20 + 0.5} = 0.24\) A

Q803. A cell of EMF 8 V and internal resistance 0.3 ohm is connected to an external resistance of 15 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{8}{15 + 0.3} = 0.52\) A

Q804. A cell of EMF 12 V and internal resistance 0.3 ohm is connected to an external resistance of 29 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{12}{29 + 0.3} = 0.41\) A

Q805. A cell of EMF 10 V and internal resistance 0.5 ohm is connected to an external resistance of 29 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{10}{29 + 0.5} = 0.34\) A

Q806. A cell of EMF 3 V and internal resistance 0.5 ohm is connected to an external resistance of 6 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{6 + 0.5} = 0.46\) A

Q807. A cell of EMF 5 V and internal resistance 0.3 ohm is connected to an external resistance of 20 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{5}{20 + 0.3} = 0.25\) A

Q808. A cell of EMF 3 V and internal resistance 0.5 ohm is connected to an external resistance of 8 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{8 + 0.5} = 0.35\) A

Q809. A cell of EMF 7 V and internal resistance 2.0 ohm is connected to an external resistance of 29 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{7}{29 + 2.0} = 0.23\) A

Q810. A cell of EMF 12 V and internal resistance 1.5 ohm is connected to an external resistance of 28 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{12}{28 + 1.5} = 0.41\) A

Q811. A cell of EMF 6 V and internal resistance 1.0 ohm is connected to an external resistance of 10 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{6}{10 + 1.0} = 0.55\) A

Q812. A cell of EMF 3 V and internal resistance 0.2 ohm is connected to an external resistance of 27 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{3}{27 + 0.2} = 0.11\) A

Q813. A cell of EMF 2 V and internal resistance 0.5 ohm is connected to an external resistance of 5 ohm. Find the current in the circuit.

Difficulty: Medium · Topic: EMF, internal resistance and current

Solution

\(I = \frac{E}{R + r} = \frac{2}{5 + 0.5} = 0.36\) A

Q814. Three resistors of 4 Ω, 6 Ω, and 12 Ω are connected in parallel. This combination is connected in series with a 2 Ω resistor and a 6 V battery. Find the total current drawn from the battery.

Difficulty: Medium-Hard · Topic: Resistors in Series and Parallel

Solution

Step 1: Find equivalent resistance of parallel combination

1/R_p = 1/4 + 1/6 + 1/12 = (3 + 2 + 1)/12 = 6/12 = 1/2

R_p = 2 Ω

Step 2: Total resistance (series with 2 Ω)

R_total = R_p + 2 = 2 + 2 = 4 Ω

Step 3: Total current

I = V/R_total = 6/4 = 1.5 A

Q815. Two lamps, one rated 60 W at 220 V and the other 100 W at 220 V, are connected in parallel to a 220 V supply. Find the total current drawn from the supply.

Difficulty: Medium-Hard · Topic: Electric Power

Solution

Step 1: Find current through each lamp

For 60 W lamp: P₁ = V × I₁ → I₁ = P₁/V = 60/220 = 3/11 A

For 100 W lamp: P₂ = V × I₂ → I₂ = P₂/V = 100/220 = 5/11 A

Step 2: In parallel, total current = sum of individual currents

I = I₁ + I₂ = 3/11 + 5/11 = 8/11 A ≈ 0.727 A

Alternative method:

Total power = 60 + 100 = 160 W

I = P/V = 160/220 = 8/11 A ✓

Q816. Compare the power consumed in the two cases: (i) a 60 W, 220 V bulb connected to a 220 V supply, and (ii) a 60 W, 220 V bulb connected to a 110 V supply.

Difficulty: Medium-Hard · Topic: Electric Power

Solution

Case (i): Bulb on 220 V supply

The bulb operates at its rated voltage, so it consumes its rated power = 60 W.

First, find the resistance of the bulb:

R = V²/P = (220)²/60 = 48400/60 = 806.67 Ω

(The resistance is a property of the filament and stays approximately the same.)

Case (ii): Bulb on 110 V supply

P = V²/R = (110)²/806.67 = 12100/806.67 = 15 W

Comparison: At half the rated voltage, the power consumed is (110/220)² = (1/2)² = 1/4 of the rated power. The bulb glows very dimly.

Q817. Find the equivalent resistance between points A and B in a circuit where a 3 Ω and 6 Ω resistor are in parallel, and this parallel combination is in series with a 4 Ω resistor.

Difficulty: Hard · Topic: Resistors in Series and Parallel

Solution

Step 1: Equivalent resistance of 3 Ω and 6 Ω in parallel

1/R_p = 1/3 + 1/6 = (2 + 1)/6 = 3/6 = 1/2

R_p = 2 Ω

Step 2: This parallel combination is in series with 4 Ω

R_eq = R_p + 4 = 2 + 4 = 6 Ω

Other Chapters in Physics

Ch 1: Light — Reflection and RefractionCh 2: The Human Eye and the Colourful WorldCh 4: Magnetic Effects of Electric CurrentCh 5: Sources of Energy

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