Chapter 10 Practice Problems — AC Power Analysis
These problems are meant to build your problem-solving intuition. A hint is provided for each — try the problem on your own first before reading it.
Problem 1 — Real, Reactive, and Apparent Power
A load is connected to a 120 V (rms), 60 Hz source. The load draws 8 A (rms) at a power factor of 0.75 lagging.
(a) Find the apparent power S.
(b) Find the real power P.
(c) Find the reactive power Q.
(d) Is the load primarily inductive or capacitive?
Hint
(a) Apparent power is simply \(S = V_{rms} \cdot I_{rms}\) in volt-amperes (VA).
(b) Real power uses the power factor: \(P = S \cdot \text{PF} = S\cos\theta\).
(c) Use the power triangle: \(Q = \sqrt{S^2 - P^2}\), or equivalently \(Q = S\sin\theta\). First find \(\theta = \cos^{-1}(0.75)\).
(d) A lagging power factor means current lags voltage — think about which passive component causes current to lag.
Problem 2 — Power in a Series RL Load
A series RL load has R = 40 Ω and \(X_L = 30\) Ω. It is connected to a 200 V (rms) source.
(a) Find the impedance magnitude |Z| and phase angle θ.
(b) Find the rms current.
(c) Calculate P, Q, and S.
(d) Express complex power \(\mathbf{S}\) in rectangular form.
Hint
(a) For a series RL, \(Z = R + jX_L\). The magnitude is \(|Z| = \sqrt{R^2 + X_L^2}\) and the angle is \(\theta = \arctan(X_L/R)\).
(b) \(I_{rms} = V_{rms}/|Z|\).
(c) \(P = I_{rms}^2 R\), \(Q = I_{rms}^2 X_L\), \(S = I_{rms}^2 |Z|\). Verify that \(S^2 = P^2 + Q^2\).
(d) Complex power is \(\mathbf{S} = P + jQ\). For an inductive load Q is positive.
Problem 3 — Power Factor Correction
An industrial load draws 10 kW at 0.60 lagging power factor from a 480 V (rms), 60 Hz supply.
(a) Find the current drawn before correction.
(b) Find the reactive power Q before correction.
(c) What value of capacitor (in μF) must be placed in parallel with the load to raise the power factor to 0.95 lagging?
(d) Find the new line current after correction.
Hint
(a) \(P = V_{rms} I_{rms} \cos\theta\). Solve for \(I_{rms}\).
(b) \(Q_{before} = P \tan\theta_{before}\), where \(\theta_{before} = \cos^{-1}(0.60)\).
(c) The capacitor must supply reactive power \(Q_C = Q_{before} - Q_{after}\), where \(Q_{after} = P\tan(\cos^{-1}(0.95))\). Then \(Q_C = V_{rms}^2 / X_C\) and \(X_C = 1/(\omega C)\). Solve for C.
(d) With the corrected power factor, re-apply \(P = V_{rms} I_{new} \cos\theta_{new}\). The real power P does not change.
Problem 4 — Maximum Power Transfer in AC
A source has Thévenin equivalent \(V_{th} = 100\angle 0°\) V and \(Z_{th} = 4 + j3\) Ω.
(a) What load impedance \(Z_L\) maximizes power delivered to the load?
(b) What is the maximum average power transferred to the load?
(c) If the load is constrained to be purely resistive, what value of \(R_L\) maximizes transfer, and what is that power?
Hint
(a) Maximum power transfer in AC occurs when the load is the complex conjugate of the source impedance: \(Z_L = Z_{th}^*\). If \(Z_{th} = R_{th} + jX_{th}\), then \(Z_L = R_{th} - jX_{th}\).
(b) At conjugate match, the reactances cancel and the circuit reduces to purely resistive. The current is \(I = V_{th}/(2R_{th})\) and \(P_{max} = \frac{|V_{th}|^2}{8R_{th}}\).
(c) For a purely resistive load, the optimal \(R_L = |Z_{th}|\). Then compute I and \(P = I^2 R_L\).
Problem 5 — Power Triangle and Efficiency
A transformer supplies three parallel loads: - Load 1: 2 kW at unity power factor - Load 2: 3 kVA at 0.80 lagging power factor - Load 3: 1.5 kVAR capacitive (purely reactive, leading)
(a) Find the total real power \(P_{total}\).
(b) Find the total reactive power \(Q_{total}\) (watch signs: inductive Q > 0, capacitive Q < 0).
(c) Find the total apparent power \(S_{total}\) and overall power factor.
Hint
(a) Real powers add directly. Convert Load 2: \(P_2 = S_2 \cdot \text{PF}_2\).
(b) Reactive powers also add algebraically. For Load 2: \(Q_2 = S_2 \sin\theta_2\) (lagging → positive). Load 3 is capacitive: \(Q_3 = -1.5\) kVAR. Load 1 has unity PF so \(Q_1 = 0\).
(c) \(S_{total} = \sqrt{P_{total}^2 + Q_{total}^2}\) and \(\text{PF} = P_{total}/S_{total}\). Note that combining an inductive and a capacitive load partially cancels reactive power, improving the overall power factor.