04-BS-4, December 2016, Q4 | Nivaar Exam Prep

Question 4 (20 marks)

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. A source $v_s(t)$ feeding a load through a series line impedance; a switched capacitor branch $X_C$ can be placed in parallel with the load. $v_s(t)=\sqrt2\,100\cos(120\pi t)$ V gives an rms phasor $V_s=100\angle0^\circ$ V at $f=60$ Hz.

Given data
RLine	2 Ω	XLine	2 Ω
RLoad	6 Ω	XLoad	4 Ω
XC	100 Ω	Vs (rms)	100 V ∠0°
f	60 Hz

Find. (a) switch open: $|I_{src}|$ and $P_{source}$; (b) switch open: $P_{line}$ and $P_{load}$; (c) switch closed: $|I_{src}|$; (d) switch closed: $P_{line}$ and $P_{load}$.

Figure 4 — line impedance feeding a load, with a switched capacitor branch XC in parallel with the load.

Approach. With the switch open the capacitor branch is absent, so it is a simple series circuit $Z_{line}+Z_{load}$. With the switch closed, $X_C$ appears in parallel with the load; find the parallel combination first, then the series total, then use a current divider to isolate the load current for the power calculation.

(a) Switch open — source current and power. $Z_{line}=2+j2\;\Omega$, $Z_{load}=6+j4\;\Omega$, so $Z_{tot}=8+j6\;\Omega$ ($|Z_{tot}|=10\;\Omega$): $$ I=\frac{V_s}{Z_{tot}}=\frac{100\angle0^\circ}{8+j6}=5.657-j4.243\text{ A}\;\Rightarrow\;|I|=\boxed{7.071\text{ A}} $$ $$ P_{source}=\operatorname{Re}(V_sI^*)=\boxed{400\text{ W}} $$
(b) Switch open — line and load power. The same current flows through both series elements: $$ P_{line}=|I|^2R_{line}=7.071^2(2)=\boxed{100\text{ W}},\qquad P_{load}=|I|^2R_{load}=7.071^2(6)=\boxed{300\text{ W}} $$ (check: $100+300=400\text{ W}=P_{source}$, confirming energy balance.)
(c) Switch closed — source current. $Z_C=-j100\;\Omega$ is now in parallel with $Z_{load}$: $$ Z_p=\frac{Z_{load}Z_C}{Z_{load}+Z_C}=\frac{(6+j4)(-j100)}{6-j96} $$ so $Z_{tot}=Z_{line}+Z_p$ and $$ I_{src}=\frac{V_s}{Z_{tot}}=5.704-j3.873\text{ A}\;\Rightarrow\;|I_{src}|=\boxed{6.894\text{ A}} $$
(d) Switch closed — line and load power. Line power uses the full source current; load power needs the load-branch current specifically, found from the node voltage $V_n=I_{src}Z_p$ and a current divider $I_{load}=V_n/Z_{load}$: $$ P_{line}=|I_{src}|^2R_{line}=6.894^2(2)=\boxed{95.07\text{ W}} $$ $$ P_{load}=|I_{load}|^2R_{load}=\boxed{308.26\text{ W}} $$ (The parallel capacitor supplies leading reactive current that partially cancels the load's lagging current, reducing $|I_{src}|$ below $|I|$ open-switch even though $P_{load}$ rises — the shunt capacitor raises the node voltage $V_n$ above the open-switch value, delivering more real power to the load while drawing less current from the source.)

Question 4 — final results
Quantity	Switch open	Switch closed
$\|I_{src}\|$	7.071 A	6.894 A
$P_{source}$	400 W	—
$P_{line}$	100 W	95.07 W
$P_{load}$	300 W	308.26 W

Topic: AC steady-state phasor analysis and real (active) power in a line-plus-load network with power-factor correction.
Key relations: $I=V/Z$ (phasor Ohm's law); $P=\operatorname{Re}(VI^*)=|I|^2R$; parallel combination $Z_p=Z_1Z_2/(Z_1+Z_2)$; current divider to recover a branch current from the total.
Why this works: A capacitor in parallel with an inductive load supplies part of the load's reactive current locally, so less reactive (and hence less total) current has to flow through the line impedance — this is the working principle of power-factor correction.
Common pitfall: Using the total source current $I_{src}$ (instead of the load-branch current $I_{load}$) to compute $P_{load}$ once the capacitor branch is present — the two currents differ once the load and capacitor form a current divider.
Source: Sadiku, Fundamentals of Electric Circuits, 7th ed., Ch. 11 (AC Power Analysis).

Question 4 of 7

Question 4 (20 marks)