Question 5 of 6: AC power — supply current, power factor, complex power

Question 5: AC power — supply current, power factor, complex power [4 + 4 + 4 + (2+3+3)]

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. $V_s=110\angle0^\circ\text{ V rms}$ feeds a $3\,\Omega$ series resistor, then two parallel branches: $(5+j10)\,\Omega$ and $(2-j4)\,\Omega$.

Find. $I_s$, the $V_s$–$I_s$ phasor diagram, the power factor, and $S$, $P$, $Q$.

[Figure not reproduced: Figure 5 — redrawn. Series $3\,\Omega$ feeds the parallel pair $Z_1=5+j10\,\Omega$ (inductive) and $Z_2=2-j4\,\Omega$ (capacitive). See the official exam paper.]

Approach. Combine the two branches, add the series $3\,\Omega$, divide $V_s$ by the total impedance for $I_s$, then form the complex power $S=V_sI_s^{*}$.

1. Parallel combination. With $Z_1=5+j10$ and $Z_2=2-j4$, $$Z_1Z_2=(5+j10)(2-j4)=50+j0,\qquad Z_1+Z_2=7+j6,$$ $$Z_p=\frac{50}{7+j6}=\frac{50(7-j6)}{85}=4.118-j3.529\,\Omega.$$
2. Total impedance. $$Z_T=3+Z_p=7.118-j3.529=7.945\angle{-26.38^\circ}\,\Omega.$$
3. (a) Supply current. $$\mathbf I_s=\frac{\mathbf V_s}{Z_T}=\frac{110\angle0^\circ}{7.945\angle{-26.38^\circ}}=\boxed{13.85\angle26.38^\circ\text{ A (rms)}}.$$ The current leads the voltage, so the net load is capacitive.
4. (b) Phasor diagram. $V_s$ lies along the reference axis; $I_s$ is rotated $+26.38^\circ$ (leading):
5. (c) Power factor. $$\text{pf}=\cos(\angle Z_T)=\cos(26.38^\circ)=\boxed{0.896\ \text{leading}}.$$
6. (d) Complex power. $$\mathbf S=\mathbf V_s\,\mathbf I_s^{*}=110\angle0^\circ\times13.85\angle{-26.38^\circ}=1523\angle{-26.38^\circ}\text{ VA}.$$ $$\boxed{S=1523\text{ VA},\quad P=1365\text{ W},\quad Q=-676\text{ VAR (capacitive)}}$$ Cross-check: $P=|I_s|^2\,\mathrm{Re}(Z_T)=13.85^2(7.118)=1365\text{ W}$ ✓.