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Chapter 10 Quiz — AC Power Analysis

Multiple Choice Questions

1. A load draws 8 A RMS from a 120 V RMS source at a phase angle of 30°. What is the real power?

  • A) 960 W
  • B) 831 W
  • C) 480 W
  • D) 1,109 W
Answer

B) 831 W

\(P = V_{rms} I_{rms}\cos\theta = 120 \times 8 \times \cos 30° = 960 \times 0.866 = 831\) W.

2. What is the unit of reactive power?

  • A) Watt (W)
  • B) Volt-Ampere (VA)
  • C) Volt-Ampere Reactive (VAR)
  • D) Joule (J)
Answer

C) Volt-Ampere Reactive (VAR)

Reactive power Q is measured in VAR (volt-ampere reactive) to distinguish it from real power (watts) and apparent power (VA).

3. A load has P = 600 W and Q = 800 VAR. What is the apparent power S?

  • A) 1,400 VA
  • B) 200 VA
  • C) 1,000 VA
  • D) 700 VA
Answer

C) 1,000 VA

\(S = \sqrt{P^2 + Q^2} = \sqrt{600^2 + 800^2} = \sqrt{360{,}000 + 640{,}000} = \sqrt{1{,}000{,}000} = 1{,}000\) VA.

4. Using the values in Q3 (P = 600 W, Q = 800 VAR, S = 1,000 VA), what is the power factor?

  • A) 0.8 lagging
  • B) 0.6 lagging
  • C) 0.8 leading
  • D) 0.6 leading
Answer

B) 0.6 lagging

\(PF = P/S = 600/1{,}000 = 0.6\). Since Q > 0, the load is inductive (lagging).

5. A pure capacitor in an AC circuit has which of the following power characteristics?

  • A) Real power only
  • B) Reactive power only, Q > 0 (absorbs VARs)
  • C) Reactive power only, Q < 0 (supplies VARs)
  • D) Equal real and reactive power
Answer

C) Reactive power only, Q < 0 (supplies VARs)

A pure capacitor has zero real power (\(P = 0\)). Current leads voltage by 90°, giving negative reactive power — the capacitor acts as a source of VARs rather than a load.

6. To improve power factor from 0.7 lagging to unity, you should add a _ in _ with the load.

  • A) Inductor, series
  • B) Capacitor, series
  • C) Resistor, parallel
  • D) Capacitor, parallel
Answer

D) Capacitor, parallel

A parallel capacitor supplies VARs locally to cancel the inductive VARs of the load, improving power factor without affecting the real power delivered to the load.

7. For maximum power transfer in an AC circuit, the load impedance should be:

  • A) Equal to the source impedance: \(Z_L = Z_s\)
  • B) The complex conjugate of source impedance: \(Z_L = Z_s^*\)
  • C) Zero (short circuit)
  • D) Much larger than source impedance
Answer

B) The complex conjugate of source impedance: \(Z_L = Z_s^*\)

Conjugate matching cancels the reactive parts (\(X_L = -X_s\)) and makes the real parts equal (\(R_L = R_s\)), maximizing power transfer.

8. At the conjugate-matched condition (maximum power transfer), what is the efficiency?

  • A) 100%
  • B) 75%
  • C) 63.2%
  • D) 50%
Answer

D) 50%

When \(R_L = R_s\), equal power is dissipated in the load and source resistance, giving 50% efficiency — the same result as in DC circuits.

9. A motor draws 2,000 VA with a power factor of 0.85 lagging. What is the reactive power?

  • A) 1,700 VAR
  • B) 1,054 VAR
  • C) 300 VAR
  • D) 2,353 VAR
Answer

B) 1,054 VAR

\(P = S \times PF = 2{,}000 \times 0.85 = 1{,}700\) W. \(Q = \sqrt{S^2 - P^2} = \sqrt{2{,}000^2 - 1{,}700^2} = \sqrt{4{,}000{,}000 - 2{,}890{,}000} = \sqrt{1{,}110{,}000} \approx 1{,}054\) VAR.

10. Power gain of an amplifier in dB is 20 dB. By what factor is the power amplified?

  • A) 20×
  • B) 10×
  • C) 100×
  • D) 40×
Answer

C) 100×

\(A_P(dB) = 10\log_{10}(P_{out}/P_{in})\), so \(20 = 10\log_{10}(A_P)\), giving \(\log_{10}(A_P) = 2\), so \(A_P = 10^2 = 100\).

Practice Problems

Problem 1: Complete Power Analysis

A 240 V RMS, 60 Hz source drives a parallel combination of: - R = 120 Ω (purely resistive) - \(X_L = 80\) Ω (purely inductive)

Find P, Q, S, and the power factor.

Solution

Power in each branch (using V = 240 V RMS across both):

Resistor: \(P_R = V^2/R = 240^2/120 = 480\) W, \(Q_R = 0\)

Inductor: \(P_L = 0\), \(Q_L = V^2/X_L = 240^2/80 = 720\) VAR

Totals:

\(P = 480\) W, \(Q = 720\) VAR

\(S = \sqrt{480^2 + 720^2} = \sqrt{230{,}400 + 518{,}400} = \sqrt{748{,}800} \approx 865\) VA

\(PF = P/S = 480/865 \approx 0.555\) lagging

Problem 2: Power Factor Correction

The circuit in Problem 1 (240 V, 60 Hz, P = 480 W, Q = 720 VAR) needs to be corrected to PF = 0.95 lagging.

a) Find the required capacitive reactive power \(Q_C\).
b) Find the required capacitance C.
c) Find the new line current after correction.

Solution

a) Required \(Q_C\):

Target: \(\theta_2 = \cos^{-1}(0.95) = 18.19°\), \(\tan\theta_2 = 0.329\)

\(Q_{new} = P\tan\theta_2 = 480 \times 0.329 = 157.9\) VAR

\(Q_C = Q_{old} - Q_{new} = 720 - 157.9 = 562.1\) VAR (capacitor must supply this)

b) Capacitance:

\(C = Q_C/(\omega V^2) = 562.1/(2\pi \times 60 \times 240^2) = 562.1/271{,}433 \approx 2.07\) μF

c) New line current:

\(S_{new} = P/PF_{new} = 480/0.95 = 505.3\) VA

\(I_{new} = S_{new}/V = 505.3/240 = 2.11\) A

Before correction: \(I_{old} = 865/240 = 3.60\) A — a 41% reduction in line current!

Problem 3: Maximum Power Transfer

A source has Thévenin equivalent \(V_s = 10\angle 0°\) V and \(Z_s = 3 + j4\) Ω.

a) What load impedance maximizes power transfer?
b) What is the maximum power delivered to the load?
c) What is the efficiency?

Solution

a) \(Z_L = Z_s^* = 3 - j4\) Ω (conjugate of source impedance)

b) At conjugate matching, reactances cancel. The circuit becomes \(V_s\) driving \(R_s + R_L = 3 + 3 = 6\) Ω:

\(I = |V_s|/(R_s + R_L) = 10/6 = 1.667\) A RMS

\(P_{max} = I^2 R_L = (1.667)^2 \times 3 = 8.33\) W

Or: \(P_{max} = |V_s|^2/(4R_s) = 100/(4 \times 3) = 8.33\) W ✓

c) \(P_{source} = I^2 \times R_s = (1.667)^2 \times 3 = 8.33\) W (same as load)

Efficiency = \(8.33/(8.33 + 8.33) = 50\%\)

Problem 4: Complex Power Calculation

A load has impedance \(Z = 6 + j8\) Ω and is connected to a 100 V RMS source.

a) Find the current phasor \(\mathbf{I}\).
b) Calculate the complex power \(\mathbf{S} = \mathbf{V}\mathbf{I}^*\).
c) Verify using \(\mathbf{S} = I_{rms}^2 Z\).
d) State P, Q, S, and PF.

Solution

a) \(|Z| = \sqrt{6^2 + 8^2} = 10\) Ω, \(\theta_Z = \tan^{-1}(8/6) = 53.13°\)

Let \(\mathbf{V} = 100\angle 0°\) V (reference).

\(\mathbf{I} = \mathbf{V}/Z = 100\angle 0° / 10\angle 53.13° = 10\angle -53.13°\) A

So \(I_{rms} = 10\) A, current lags voltage by 53.13°.

b) \(\mathbf{I}^* = 10\angle +53.13° = 10(0.6 + j0.8) = 6 + j8\) A

\(\mathbf{S} = \mathbf{V}\mathbf{I}^* = 100\angle 0° \times (6+j8) = 600 + j800\) VA

c) \(\mathbf{S} = I^2 Z = 100 \times (6+j8) = 600 + j800\) VA ✓

d) \(P = 600\) W, \(Q = 800\) VAR (lagging), \(S = \sqrt{600^2+800^2} = 1{,}000\) VA

\(PF = P/S = 0.6\) lagging