Question 5 of 7: AC Power Flow with a Switched Power-Factor Capacitor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2015 — 04-BS-4 Electric Circuits and Power. Closed book; one aid sheet (both sides) and a Casio/Sharp approved calculator permitted; 3 hours. Seven questions; any five constitute a complete paper (only the first five in the answer book are marked). All seven are solved here as a study resource. Marks: each question 20 (four 5-mark parts).
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (KCL/KVL, Thévenin, first-order transients, AC power); R. Boylestad & L. Nashelsky, Electronic Devices and Circuit Theory (bridge rectifier, RC filter); S. J. Chapman, Electric Machinery Fundamentals (magnetic circuits); M. M. Mano & M. Ciletti, Digital Design (combinational logic). Canadian frame: 60 Hz mains, SI units.
Question 5: AC Power Flow with a Switched Power-Factor Capacitor (20 marks)
Given. A source feeds an inductive load through a line impedance; a capacitor Xc can be switched in parallel with the load for power-factor correction. The source is $V_s(t)=\sqrt2\,(100)\cos(120\pi t)$ V, i.e. 100 V rms at 60 Hz.
Given data
Z_Line
Z_Load
Xc
Vs (rms)
f
2 + j2 Ω
6 + j4 Ω
−j100 Ω
100 V
60 Hz
Find. Source-current magnitude and real powers (source, line, load) for both switch states.
Figure 5 — inductive load (X_Load + R_Load) fed through the line impedance (X_Line + R_Line); the switch places the capacitor Xc in parallel with the load bus for power-factor correction.
Approach. Series impedances add; the switched capacitor sits in parallel with the load. Use $|I|=|V|/|Z|$ and $P=|I|^2R$ for each element; the source real power is $\mathrm{Re}\{V_s I_s^{*}\}$.
Switch OPEN (a, b)
a) Source current and power. $Z_{tot}=Z_{Line}+Z_{Load}=(2+j2)+(6+j4)=8+j6\ \Omega$, $|Z|=10\ \Omega$.
$$I_s=\frac{V_s}{Z_{tot}}=\frac{100}{8+j6}=8-j6\text{ A},\qquad |I_s|=10\text{ A}.$$
$$\boxed{|I_s| = 10\text{ A},\qquad P_{src}=|I_s|^2(R_{Line}+R_{Load})=100(8)=800\text{ W}}$$
b) Split between line and load. The same 10 A flows through both:
$$P_{Line}=|I_s|^2R_{Line}=100(2)=200\text{ W},\qquad P_{Load}=|I_s|^2R_{Load}=100(6)=600\text{ W}.$$
$$\boxed{P_{Line}=200\text{ W},\quad P_{Load}=600\text{ W}}$$
Switch CLOSED (c, d)
c) Source current. The capacitor parallels the load: $Z_{p}=Z_{Load}\,\|\,(-jX_C)=\dfrac{(6+j4)(-j100)}{(6+j4)-j100}=6.485+j3.761\ \Omega$. Then $Z_{tot}=Z_{Line}+Z_p=8.485+j5.761\ \Omega$, $|Z_{tot}|=10.26\ \Omega$:
$$I_s=\frac{100}{8.485+j5.761},\qquad \boxed{|I_s| = 9.75\text{ A}}$$
The capacitor has cut the source current from 10 A to 9.75 A by supplying part of the load's reactive demand.
d) Line and load real power. Line power uses the (reduced) source current; load power uses the load-branch current $I_{Load}=V_{bus}/Z_{Load}$ where $V_{bus}=I_sZ_p$:
$$P_{Line}=|I_s|^2R_{Line}=(9.75)^2(2)=190.1\text{ W},$$
$$P_{Load}=|I_{Load}|^2R_{Load}=(10.14)^2(6)=616.5\text{ W}.$$
$$\boxed{P_{Line}=190\text{ W},\quad P_{Load}=617\text{ W}}$$
The capacitor raises the load-bus voltage (less drop across the line reactance), so the fixed-impedance load draws slightly more real power while the line loss falls.
