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Current Electricity MCQs — Physics (page 8)

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.

176. A charge of 0.5 C moves through a p.d. of 230 V. The energy transferred is:

  1. 230 J
  2. 460 J
  3. 115 J
  4. 57.5 J

Answer: C) 115 J

Explanation: W = QV = 115 J. SLO P-09-E-11 · Application

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177. Resistors of 4 Ω, 4 Ω, 4 Ω, 4 Ω are connected in series. The combined resistance is:

  1. 32 Ω
  2. 160 Ω
  3. 8 Ω
  4. 16 Ω

Answer: D) 16 Ω

Explanation: R = R1 + R2 + ldots = 16 Ω. SLO P-09-E-20 · Application

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178. A wire has resistivity 1.1 10^-6 Ω m, length 0.5 m and cross-section 2 10^-7 m^2. Its resistance is:

  1. 2.75 Ω
  2. 0.275 Ω
  3. 27.5 Ω
  4. 5.5 Ω

Answer: A) 2.75 Ω

Explanation: R = rho LA = 2.75 Ω. SLO P-09-E-13 · Application

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179. At a junction, currents of 2.5 A flow in. Currents of 1.2 A, 0.3 A flow out, together with one unknown current I. I is:

  1. 2 A
  2. 0.5 A
  3. 1 A
  4. 10 A

Answer: C) 1 A

Explanation: sum Iin = sum Iout Rightarrow I = 2.5 - 1.5 = 1 A. SLO P-09-E-18 · Knowledge

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180. A 6 V battery is connected across three 2 Ω resistors in parallel. The battery current is:

  1. 0.67 A
  2. 9 A
  3. 1 A
  4. 3 A

Answer: B) 9 A

Explanation: Each branch 3 A. SLO P-09-E-22 · Application

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181. A potentiometer wire is 1 m long with 2 V across it. The potential gradient is:

  1. 200 V/m
  2. 2 V/cm
  3. 0.5 V/m
  4. 2 V/m

Answer: D) 2 V/m

Explanation: k = V/L. SLO P-09-E-23 · Application

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182. A 12 V cell with negligible internal resistance is connected to a 4 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω. The current from the cell is:

  1. 2 A
  2. 3 A
  3. 1 A
  4. 0.92 A

Answer: A) 2 A

Explanation: Parallel pair = 2 Ω; total = 6 Ω; I = 12/6. SLO P-09-E-22 · Application

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183. A circuit has a 9 V cell, internal resistance 1 Ω, and an external 8 Ω resistor. The current is:

  1. 1 A
  2. 9 A
  3. 0.89 A
  4. 1.125 A

Answer: A) 1 A

Explanation: I = 98 + 1. SLO P-09-E-22 · Application

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184. When a car's starter motor is operated, the headlights dim because:

  1. the headlights are disconnected
  2. the motor produces light
  3. the battery's e.m.f. increases
  4. the large current causes a big p.d. across the battery's internal resistance

Answer: D) the large current causes a big p.d. across the battery's internal resistance

Explanation: Terminal p.d. drops. SLO P-09-E-26 · Understanding

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185. A cell of e.m.f. 1.5 V and internal resistance 0.5 Ω is connected to a 2.5 Ω resistor. The terminal p.d. is:

  1. 0.625 V
  2. 1.25 V
  3. 2.5 V
  4. 12.5 V

Answer: B) 1.25 V

Explanation: I = mathcalER + r = 0.5 A; V = IR = 1.25 V (lost volts Ir = 0.25 V). SLO P-09-E-17 · Understanding

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186. Adding another resistor in series to a circuit makes the current:

  1. increase
  2. stay the same
  3. decrease
  4. reverse

Answer: C) decrease

Explanation: Total resistance increases. SLO P-09-E-20 · Application

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187. Ordinary concrete without fibres is a:

  1. poor conductor of electricity
  2. good conductor
  3. magnet
  4. superconductor

Answer: A) poor conductor of electricity

Explanation: Fibres provide the conducting path. SLO P-09-E-27 · Understanding

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188. A 3 Ω resistor is connected to 9 V. The power is:

  1. 27 W
  2. 28 W
  3. 3 W
  4. 54 W

Answer: A) 27 W

Explanation: P = V^2R = 9^23 = 27 W. SLO P-09-E-12 · Application

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189. Measuring resistance to detect damage is useful because it:

  1. needs the bridge to be closed for months
  2. is quick and doesn't need the bridge to be broken open
  3. requires removing the concrete
  4. needs X-ray equipment

Answer: B) is quick and doesn't need the bridge to be broken open

Explanation: Non-destructive testing. SLO P-09-E-27 · Understanding

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190. A current of 10 A flows through a 0.5 Ω resistor. The power dissipated is:

  1. 100 W
  2. 5 W
  3. 25 W
  4. 50 W

Answer: D) 50 W

Explanation: P = I^2R = (10)^2(0.5) = 50 W. SLO P-09-E-12 · Application

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191. A generator produces e.m.f. by converting:

  1. mechanical energy into electrical energy
  2. electrical into mechanical
  3. heat into chemical energy
  4. light into sound

Answer: A) mechanical energy into electrical energy

Explanation: Energy input to charges. SLO P-09-E-16 · Understanding

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192. Embedding sensors in structures for continuous checking is called:

  1. power generation
  2. radio broadcasting
  3. structural health monitoring
  4. electroplating

Answer: C) structural health monitoring

Explanation: Engineering application. SLO P-09-E-27 · Understanding

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193. In I = Anvq, n stands for:

  1. the number of turns
  2. the number of electrons in the wire
  3. the refractive index
  4. the number of charge carriers per unit volume

Answer: D) the number of charge carriers per unit volume

Explanation: Units m^-3. SLO P-09-E-10 · Application

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

  1. 0.4 A
  2. 1.2 A
  3. 2 A
  4. 0.8 A

Answer: A) 0.4 A

Explanation: Net e.m.f. = 2 V; I = 2/5. SLO P-09-E-22 · Application

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195. Two identical cells in parallel compared with one cell give:

  1. the same e.m.f. but half the internal resistance
  2. double the e.m.f.
  3. double the internal resistance
  4. zero e.m.f.

Answer: A) the same e.m.f. but half the internal resistance

Explanation: Parallel internal resistances. SLO P-09-E-26 · Understanding

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196. The potential difference between two points is:

  1. the current per unit time
  2. the energy transferred per unit charge passing between them
  3. the charge per unit energy
  4. the force per unit charge

Answer: B) the energy transferred per unit charge passing between them

Explanation: V = W/Q. SLO P-09-E-11 · Application

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197. A 24 V supply drives current through 3 Ω and R in series; the current is 2 A. R is:

  1. 12 Ω
  2. 9 Ω
  3. 6 Ω
  4. 3 Ω

Answer: B) 9 Ω

Explanation: Total = 12 Ω. SLO P-09-E-22 · Application

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198. A current of 5 A flows in a wire with n = 8.5 10^28 m^-3 and area 3 10^-6 m^2. The drift velocity of the electrons is:

  1. 1.23 10^-4 m/s
  2. 2.45 10^-4 m/s
  3. 0.00123 m/s
  4. 1.23 10^-5 m/s

Answer: A) 1.23 10^-4 m/s

Explanation: v = InAq = 1.23 10^-4 m/s — very slow. SLO P-09-E-10 · Application

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199. Taking currents into a junction as positive, Kirchhoff's first law can be written as:

  1. sum I = 0
  2. sum IR = sum mathcalE
  3. sum I = 1
  4. sum V = 0

Answer: A) sum I = 0

Explanation: Algebraic sum. SLO P-09-E-18 · Knowledge

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200. Which instrument uses a galvanometer in a null method?

  1. a thermometer
  2. a voltmeter across a lamp
  3. a potentiometer
  4. an ammeter in series

Answer: C) a potentiometer

Explanation: Balance with zero current. SLO P-09-E-24 · Understanding

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