MDCAT Physics chapter 12: Physics · Alternating current MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

19 exam-style questions on Physics · Alternating current, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

19 questions · 2 topics · 10 per test

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What chapter 12 covers

Ten Physics · Alternating current questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. A 60 Ω resistor and an inductor of reactance 80 Ω are connected in series across a 200 V rms AC supply. The average power dissipated in the circuit is

  1. A
    400 W
  2. B
    240 W
  3. C
    320 W
  4. D
    667 W
Show answer

Answer: B. Z = √(602 + 802) = 100 Ω, so Irms = 200/100 = 2.0 A. Only the resistor dissipates power: P = I2R = 4 × 60 = 240 W (the same as VI cos θ = 200 × 2.0 × 0.6). 400 W leaves out the power factor, and 320 W wrongly puts the power in the inductor.

2. A coil of negligible resistance and inductance 0.40/π H is connected to a 50 Hz supply. Its reactance is

  1. A
    20 Ω
  2. B
    40 Ω
  3. C
    6.4 Ω
  4. D
    0.025 Ω
Show answer

Answer: B. XL = 2πfL = 2π × 50 × 0.40/π = 40 Ω. Leaving out the 2 gives 20 Ω, and 0.025 Ω is 1/(2πfL), the capacitor formula applied to an inductor.

3. The voltage of an AC source is V = 200 sin (100πt) volts, with t in seconds. The frequency of the supply is

  1. A
    100 Hz
  2. B
    314 Hz
  3. C
    50 Hz
  4. D
    200 Hz
Show answer

Answer: C. Comparing with V = V0 sin ωt gives ω = 100π rad s−1, so f = ω/2π = 50 Hz. 314 is ω itself (100π ≈ 314), and 200 is the peak voltage in volts, not a frequency.

4. In a series RL circuit on an AC supply, the rms voltage across the resistor is 30 V and across the pure inductor is 40 V. The rms supply voltage is

  1. A
    70 V
  2. B
    50 V
  3. C
    10 V
  4. D
    35 V
Show answer

Answer: B. VR is in phase with the current while VL leads it by 90°, so they add as phasors: V = √(302 + 402) = 50 V. 70 V adds them as if they were in phase.

5. A series circuit has R = 40 Ω, XL = 70 Ω and XC = 40 Ω. Its impedance is

  1. A
    150 Ω
  2. B
    50 Ω
  3. C
    90 Ω
  4. D
    70 Ω
Show answer

Answer: B. The voltages across L and C are 180° out of phase, so the net reactance is 70 − 40 = 30 Ω and Z = √(402 + 302) = 50 Ω. 90 Ω squares and adds all three values, and 150 Ω simply adds them.

6. A 12 Ω resistor is in series with a capacitor of reactance 5 Ω across an AC supply. The impedance of the circuit is

  1. A
    13 Ω
  2. B
    17 Ω
  3. C
    7 Ω
  4. D
    169 Ω
Show answer

Answer: A. The voltages across R and C are 90° out of phase, so Z = √(R2 + XC2) = √(144 + 25) = √169 = 13 Ω. 17 Ω adds the two as if they were in phase, and 169 Ω forgets the square root.

7. A capacitor has a reactance of 120 Ω at 50 Hz. Its reactance at 150 Hz is

  1. A
    360 Ω
  2. B
    120 Ω
  3. C
    40 Ω
  4. D
    13.3 Ω
Show answer

Answer: C. XC = 1/(2πfC) is inversely proportional to f, so tripling the frequency divides the reactance by 3: 120/3 = 40 Ω. 360 Ω treats the capacitor like an inductor, whose reactance rises with f.

8. An alternating current I = I0 sin ωt starts from zero at t = 0. At what phase angle ωt does the instantaneous current first become equal to the rms value?

  1. A
    45°
  2. B
    30°
  3. C
    60°
  4. D
    90°
Show answer

Answer: A. I equals Irms = I0/√2 when sin ωt = 1/√2, that is ωt = 45° (π/4 rad). At 30° the current is only I0/2, and at 90° it is at its peak I0.

9. An oscilloscope shows a sinusoidal voltage with a peak-to-peak value of 20√2 V. Its rms value is

  1. A
    20 V
  2. B
    10√2 V
  3. C
    10 V
  4. D
    40 V
Show answer

Answer: C. The peak value is half the peak-to-peak value, V0 = 10√2 V, and Vrms = V0/√2 = 10 V. 20 V divides the peak-to-peak value by √2, and 10√2 V is the peak value itself.

10. In an AC circuit containing only a capacitor, the current

  1. A
    leads the voltage by π/2
  2. B
    lags the voltage by π/2
  3. C
    is in phase with the voltage
  4. D
    leads the voltage by π
Show answer

Answer: A. The current is greatest when the voltage is changing fastest (as it passes through zero), so the current reaches its peak a quarter cycle before the voltage: it leads by π/2 (90°). In a pure inductor the current lags by π/2.

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Physics · Alternating current: questions students ask

How many Physics · Alternating current MCQs are there?

19, each with the answer and a short explanation, across 2 topics: AC quantities, AC circuits. Every chapter test draws ten of them, so you can take it many times.

Is the MDCAT Physics chapter 12 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

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