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Current Electricity MCQs — Physics

270 multiple-choice questions on Current Electricity from Physics, each with the correct answer and, where available, a teacher's explanation. Read through them to revise the chapter, then practise the same questions one by one to test yourself.

1. A voltmeter across a cell reads 1.5 V on open circuit and 1.2 V when a lamp is connected. The 'lost volts' are:

  1. 1.2 V
  2. 1.5 V
  3. 2.7 V
  4. 0.3 V

Answer: D) 0.3 V

Explanation: 1.5 - 1.2. SLO P-09-E-26 · Understanding

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2. A wire of length 2 m and diameter 0.5 mm has resistance 5 Ω. Its resistivity is:

  1. 4.91 10^-6 Ω m
  2. 1.96 10^-6 Ω m
  3. 1.23 10^-7 Ω m
  4. 4.91 10^-7 Ω m

Answer: D) 4.91 10^-7 Ω m

Explanation: A = pi r^2 = 1.96 10^-7 m^2; rho = RAL = 4.91 10^-7 Ω m. SLO P-09-E-13 · Application

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3. A 60 W, 230 V lamp has a resistance of about:

  1. 882 Ω
  2. 13,800 Ω
  3. 3.8 Ω
  4. 0.26 Ω

Answer: A) 882 Ω

Explanation: R = V^2P = 52,90060. SLO P-09-E-12 · Application

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4. A 6 V battery (no internal resistance) is connected to three resistors in series. The p.d.s across two of them are 1.2 V and 2.8 V. The p.d. across the third is:

  1. 20 V
  2. 2 V
  3. 4 V
  4. 1 V

Answer: B) 2 V

Explanation: Kirchhoff's second law: sum V = mathcalE Rightarrow V3 = 6 - 1.2 - 2.8 = 2 V. SLO P-09-E-19 · Understanding

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5. Car batteries need a very low internal resistance because:

  1. they are rechargeable
  2. they work at high temperature
  3. the starter motor draws a very large current
  4. they are heavy

Answer: C) the starter motor draws a very large current

Explanation: Large current without big lost volts. SLO P-09-E-17 · Understanding

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6. A cell of e.m.f. 9 V and internal resistance 2 Ω is connected to a 7 Ω resistor. The terminal p.d. is:

  1. 7 V
  2. 14 V
  3. 3.5 V
  4. 70 V

Answer: A) 7 V

Explanation: I = mathcalER + r = 1 A; V = IR = 7 V (lost volts Ir = 2 V). SLO P-09-E-17 · Understanding

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7. 1 volt is equal to:

  1. 1 coulomb per joule
  2. 1 ampere per second
  3. 1 watt per ampere squared
  4. 1 joule per coulomb

Answer: D) 1 joule per coulomb

Explanation: J/C. SLO P-09-E-11 · Application

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8. A wire has resistivity 2.8 10^-8 Ω m, length 100 m and cross-section 2.5 10^-6 m^2. Its resistance is:

  1. 11.2 Ω
  2. 0.112 Ω
  3. 2.24 Ω
  4. 1.12 Ω

Answer: D) 1.12 Ω

Explanation: R = rho LA = 1.12 Ω. SLO P-09-E-13 · Application

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9. A potential divider has 1 kΩ and 2 kΩ resistors in series across 6 V. The p.d. across the 2 kΩ resistor is:

  1. 4 V
  2. 40 V
  3. 8 V
  4. 2 V

Answer: A) 4 V

Explanation: Vout = Vin R2R1 + R2 = 6 23 = 4 V. SLO P-09-E-25 · Understanding

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10. Two cells, 6 V and 4 V, are connected in series aiding with a 5 Ω resistor. The current is:

  1. 2 A
  2. 10 A
  3. 0.4 A
  4. 1.2 A

Answer: A) 2 A

Explanation: Net e.m.f. = 10 V; I = 10/5. SLO P-09-E-22 · Application

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11. The charge carriers in a metal are:

  1. free electrons
  2. neutrons
  3. positive ions moving freely
  4. protons

Answer: A) free electrons

Explanation: Delocalised electrons. SLO P-09-E-10 · Application

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12. Potential difference is measured with a voltmeter connected:

  1. anywhere in the circuit
  2. in parallel with the component
  3. across the battery only
  4. in series with the component

Answer: B) in parallel with the component

Explanation: Measures energy per charge across it. SLO P-09-E-11 · Application

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13. Resistors of 6 Ω, 3 Ω are connected in parallel. The combined resistance is:

  1. 4 Ω
  2. 20 Ω
  3. 1 Ω
  4. 2 Ω

Answer: D) 2 Ω

Explanation: 1R = 16 + 13 Rightarrow R = 2 Ω. SLO P-09-E-21 · Application

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14. The galvanometer in a potentiometer circuit is connected:

  1. nowhere
  2. in parallel with the whole wire
  3. across the driver cell only
  4. in series with the cell being tested and the jockey

Answer: D) in series with the cell being tested and the jockey

Explanation: Detects current from the test cell. SLO P-09-E-24 · Understanding

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15. A potential divider has 4 kΩ and 8 kΩ resistors in series across 12 V. The p.d. across the 8 kΩ resistor is:

  1. 8 V
  2. 80 V
  3. 16 V
  4. 4 V

Answer: A) 8 V

Explanation: Vout = Vin R2R1 + R2 = 12 812 = 8 V. SLO P-09-E-25 · Understanding

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16. For a light-activated street lamp switch, the output must rise when it gets dark. The output should be taken across:

  1. the supply
  2. nothing
  3. the LDR
  4. the fixed resistor

Answer: C) the LDR

Explanation: Its share of p.d. rises in the dark. SLO P-09-E-25 · Understanding

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17. A 12 V source with internal resistance 2 Ω supplies a 4 Ω resistor. The efficiency of the power transfer is:

  1. 33.3%
  2. 66.7%
  3. 50%
  4. 100%

Answer: B) 66.7%

Explanation: Efficiency = I^2RI^2(R + r) = RR + r = 66.7%. SLO P-09-E-26 · Understanding

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18. In a series circuit, Kirchhoff's first law means the current is:

  1. different in each component
  2. the same at every point
  3. zero after each component
  4. largest near the battery

Answer: B) the same at every point

Explanation: No junctions. SLO P-09-E-18 · Knowledge

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19. For two resistors in parallel, R =

  1. R1 + R2
  2. R1 + R2R1R2
  3. R1R2
  4. R1R2R1 + R2

Answer: D) R1R2R1 + R2

Explanation: Product over sum. SLO P-09-E-21 · Application

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20. In a potential divider, an LDR is in series with a fixed resistor; the output is taken across the LDR. When it gets darker, the output:

  1. decreases
  2. stays the same
  3. becomes zero
  4. increases

Answer: D) increases

Explanation: LDR resistance rises, so it takes a larger share of the p.d. SLO P-09-E-25 · Understanding

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21. High-voltage power supplies (EHT) have a very high internal resistance to:

  1. increase the current
  2. limit the current for safety
  3. save money
  4. increase the e.m.f.

Answer: B) limit the current for safety

Explanation: Protects users. SLO P-09-E-17 · Understanding

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22. 100 J of energy is transferred when 5 C of charge passes between two points. The p.d. is:

  1. 500 V
  2. 20 V
  3. 40 V
  4. 10 V

Answer: B) 20 V

Explanation: V = WQ = 1005 = 20 V. SLO P-09-E-11 · Application

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23. A device draws 2 A from a 12 V supply. Its power is:

  1. 48 W
  2. 24 W
  3. 12 W
  4. 6 W

Answer: B) 24 W

Explanation: P = IV = 24 W. SLO P-09-E-12 · Application

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24. Resistors of 4 Ω, 4 Ω are connected in parallel. The combined resistance is:

  1. 1 Ω
  2. 4 Ω
  3. 2 Ω
  4. 20 Ω

Answer: C) 2 Ω

Explanation: 1R = 14 + 14 Rightarrow R = 2 Ω. SLO P-09-E-21 · Application

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25. A solar cell's e.m.f. comes from converting:

  1. nuclear energy
  2. chemical energy
  3. kinetic energy
  4. light energy into electrical energy

Answer: D) light energy into electrical energy

Explanation: Photovoltaic effect. SLO P-09-E-15 · Application

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