AC power and power-factor correction: 22-Elec-A1 Circuits, May 2013, Q5

Question 5: AC power and power-factor correction [4 + 4 + 4 + 4 + 4]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $V_s=120\angle 0^\circ$ V rms, $f=60\ \text{Hz}$; Load-1 $P_1=5\ \text{kW}$ (unity pf); Load-2 $S_2=10\ \text{kVA}$ at $\text{pf}=0.6$ lagging.

Find. $I_1,I_2,I_s$, the source pf, $S_s$, and the correction capacitance $C$.

Figure 6 — Parallel household loads on the 120 V supply, with the power-factor-correction capacitor branch (initially open).

Approach. Convert each load to a phasor current (magnitude $S/V$, angle set by its power factor), add for the supply current, form the complex power, and size the capacitor from the reactive-power deficit.

Load currents. Load-1 is resistive: $I_1=P_1/V=5000/120=41.67\ \text{A}$ at $0^\circ$. Load-2: $|I_2|=S_2/V=10000/120=83.33\ \text{A}$ at $-\cos^{-1}0.6=-53.13^\circ$: $$\boxed{I_1=41.67\angle 0^\circ\ \text{A},\qquad I_2=83.33\angle{-53.13^\circ}\ \text{A}.}$$
Supply current. Adding rectangular parts, $I_1=41.67$, $I_2=50-j66.67$: $$I_s=I_1+I_2=91.67-j66.67=\boxed{113.35\angle{-36.03^\circ}\ \text{A}.}$$
Source power factor. $$\text{pf}=\cos(36.03^\circ)=\boxed{0.809\ \text{lagging}.}$$
Source complex power. $S_s=V_sI_s^{*}$, or by summing loads $P=5000+10000(0.6)=11\,000\ \text{W}$, $Q=10000(0.8)=8000\ \text{VAR}$: $$\boxed{S_s=11\,000+j8000\ \text{VA}=13.60\angle 36.03^\circ\ \text{kVA}.}$$
Correction capacitor. Target $\text{pf}=0.95$ lag ⇒ $\theta'=18.19^\circ$, $Q'=P\tan\theta'=11000(0.3287)=3616\ \text{VAR}$. The capacitor supplies $Q_C=Q-Q'=8000-3616=4384\ \text{VAR}$: $$C=\frac{Q_C}{V^{2}\,\omega}=\frac{4384}{(120)^2(2\pi\cdot 60)}=\boxed{808\ \mu\text{F}.}$$

Quantity	Result
Load currents	$I_1=41.67\angle 0^\circ$, $I_2=83.33\angle{-53.13^\circ}$ A
Supply current	$I_s=113.35\angle{-36.03^\circ}$ A
Source power factor	$0.809$ lagging
Source complex power	$S_s=11\,000+j8000$ VA $=13.60\ \text{kVA}\angle 36.03^\circ$
Correction capacitor	$\boxed{C=808\ \mu\text{F}}$

Concept & Theory

Topic. Complex power in parallel AC loads and shunt-capacitor power-factor correction.

Key relations. $S=P+jQ=VI^{*}$; $\text{pf}=\cos\theta=P/|S|$; for a load $|I|=S/V$ at angle $-\theta$ (lagging); shunt correction $Q_C=P(\tan\theta_{old}-\tan\theta_{new})$, $C=Q_C/(\omega V^2)$.

Why this works. Real and reactive powers add arithmetically across parallel loads because they share the same voltage; the capacitor injects leading VARs that cancel part of the inductive $Q$, rotating $S$ toward the real axis without changing $P$. Only the reactive component must be adjusted to hit the new power factor.

Common pitfall. Correcting to $Q=0$ (unity) instead of the target $Q'=P\tan\theta'$, or dropping the rms-versus-peak distinction when sizing $C$ (here $V$ is already rms).

Source. Sadiku, Fundamentals of Electric Circuits §11 (AC power analysis; complex power, power-factor correction).

Question 5 of 6: AC power and power-factor correction

Question 5: AC power and power-factor correction [4 + 4 + 4 + 4 + 4]