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AC circuits: board exam questions with solutions

On the ECE exam under Electronics Engineering and the REE exam under Electrical Engineering Professional Subjects and the RME exam under Technical Subjects. Try each question before you open its solution, then practise with fresh numbers below: 5 computed question types on this topic.

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RLC series circuit · RLC parallel circuit · Resonant tank (LC) · RC low-pass filter: the same circuits on a live bench, with meters you can move.

Questions

1. What is the reactance of a 10 µF capacitor at 60 Hz?

  1. 3.77 Ω
  2. 1666.7 Ω
  3. 2652.6 Ω
  4. 265.3 Ω
Show the answer and solution

Answer: D. 265.3 Ω

Xc = 1/(2πfC) = 1/(2π × 60 × 10 × 10⁻⁶) = 265.3 Ω.

2. A coil with R = 30 Ω and XL = 40 Ω is connected to 230 V, 60 Hz. What real power does it draw?

  1. 1058.0 W
  2. 846.4 W
  3. 634.8 W
  4. 1763.3 W
Show the answer and solution

Answer: C. 634.8 W

|Z| = √(30² + 40²) = 50 Ω, I = 230/50 = 4.6 A, P = I²R = 4.6² × 30 = 634.8 W.

3. What is the resonant frequency of a series circuit with L = 10 mH and C = 100 nF?

  1. 31.62 kHz
  2. 5.03 kHz
  3. 50.3 kHz
  4. 503 Hz
Show the answer and solution

Answer: B. 5.03 kHz

f0 = 1/(2π√(LC)) = 1/(2π√(10 × 10⁻³ × 100 × 10⁻⁹)) = 5033 Hz ≈ 5.03 kHz. (31.62 kHz is ω0 in rad/s ÷ 1000, a common slip.)

4. A series RLC circuit has R = 10 Ω, L = 10 mH and C = 100 nF. What is its −3 dB bandwidth?

  1. 159.2 Hz
  2. 1000 Hz
  3. 31.6 Hz
  4. 5033 Hz
Show the answer and solution

Answer: A. 159.2 Hz

BW = R/(2πL) = 10/(2π × 0.01) = 159.2 Hz. Check: f0 = 5033 Hz, Q = (1/R)√(L/C) = 31.6, f0/Q = 159.2 Hz.

5. A 10 kW load operates at 0.70 lagging power factor. How much capacitive kVAR is needed to raise the power factor to 0.95 lagging?

  1. 10.20 kVAR
  2. 2.50 kVAR
  3. 3.29 kVAR
  4. 6.92 kVAR
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Answer: D. 6.92 kVAR

Qc = P(tan θ1 − tan θ2) = 10(tan(cos⁻¹0.70) − tan(cos⁻¹0.95)) = 10(1.0202 − 0.3287) = 6.92 kVAR.

6. A balanced three-phase load draws 20 A line current from a 400 V (line-to-line) supply at 0.85 power factor. What is the total real power?

  1. 20.40 kW
  2. 6.80 kW
  3. 11.78 kW
  4. 13.86 kW
Show the answer and solution

Answer: C. 11.78 kW

P = √3 VL IL cos θ = √3 × 400 × 20 × 0.85 = 11778 W ≈ 11.78 kW.

7. In a purely capacitive AC circuit, how is the current related to the applied voltage?

  1. Current lags voltage by 90°
  2. Current leads voltage by 90°
  3. Current is in phase with voltage
  4. Current leads voltage by 180°
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Answer: B. Current leads voltage by 90°

For a capacitor i = C dv/dt; with a sinusoidal voltage the current is a cosine, leading the voltage by 90° (ICE).

8. Two loads share a 230 V, 60 Hz supply: a 5 kW resistive heater and a motor drawing 4 kVA at 0.60 lagging power factor. What is the combined power factor?

  1. 0.918 lagging
  2. 0.800 lagging
  3. 0.600 lagging
  4. 0.942 lagging
Show the answer and solution

Answer: A. 0.918 lagging

P = 5 + 4 × 0.6 = 7.4 kW, Q = 4 × 0.8 = 3.2 kVAR, S = √(7.4² + 3.2²) = 8.062 kVA, pf = 7.4/8.062 = 0.918 lagging.

Worked examples with computed numbers

These numbers are fixed on this page; in practice they change every time.

9. What is the reactance of a 10 µF capacitor at 1 kHz?

  1. 15.9 Ω
  2. 100 Ω
  3. 62.8 mΩ
  4. 15.9 mΩ
Show the answer and solution

Answer: A. 15.9 Ω

Xc = 1/(2πfC) = 1/(2π × 1000 × 10 µF) = 15.9 Ω.

10. At what frequency does 10 mH resonate with 1 nF?

  1. 50.3 kHz
  2. 15.9 GHz
  3. 316 kHz
  4. 19.9 MHz
Show the answer and solution

Answer: A. 50.3 kHz

f₀ = 1/(2π√(LC)) = 50.3 kHz.

11. A 80 Ω resistor is in series with a 50 mH inductor on 60 Hz. What is the power factor?

  1. 0.809
  2. 4.24
  3. 0.229
  4. 0.973
Show the answer and solution

Answer: D. 0.973

XL = 2π(60)(0.05) = 18.85 Ω; Z = √(R² + XL²) = 82.19 Ω; pf = R/Z = 0.973 lagging.

12. What is the Q of a series RLC circuit with R = 2 Ω, L = 10 mH and C = 100 nF?

  1. 50000
  2. 0.00632
  3. 0.0000158
  4. 158
Show the answer and solution

Answer: D. 158

Q = (1/R)√(L/C) = √(10 mH/100 nF)/2 = 158.

13. What is the −3 dB cutoff of an RC low-pass filter with R = 15 kΩ and C = 100 nF?

  1. 53.1 Hz
  2. 1.5 mHz
  3. 106 Hz
  4. 667 Hz
Show the answer and solution

Answer: C. 106 Hz

fc = 1/(2πRC) = 106 Hz; there the output is 3 dB down.

Other topics in the same subjects

DC circuits · Transients (RC, RL, RLC) · Semiconductors and diodes · BJT and FET circuits · Operational amplifiers · Oscillators and feedback · Power electronics · Digital electronics · Microprocessors and microcontrollers · Control systems · Instrumentation and measurement · Three-phase systems · Transformers · DC machines · Induction motors · Synchronous machines · Power systems and transmission · Protection: fuses, breakers, relays · Illumination · Power plant engineering · PEC: branch circuits, ampacity and protection · Tools and test equipment · Motors, starters and control circuits · RA 7920 and ethics (electrical)

An independent, free reviewer: not affiliated with the PRC or the Boards. Subject weights and passing rules are from the PRC examination programs and the laws (RA 9292, RA 7920); check the current PRC program before you file. Code questions follow the Philippine Electrical Code 2017 Part 1; check them against your own copy of the Code.