Superposition in an AC parallel network: 23-Mechatronics-A2 Circuits and Electronics, December 2019, Q5

Question 5: Superposition in an AC parallel network [20]

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.

Quantity	Value
Current source	$i_s=5\cos(20t+10^\circ)$ A
Resistor	$10\,\Omega$
Capacitor	$0.5$ F
Inductor	$2$ H
Voltage source, $V_o$ measured across $C$	$V_s=10\sin(20t)$ V

Find. $V_o(t)$ by superposition.

Figure-5: all elements in parallel between the same two rails — current source $i_s$, $10\,\Omega$, $0.5$ F capacitor ($V_o$ across it), $2$ H inductor, and voltage source $V_s$ (+ at top).

Approach. Apply each source alone (the other killed) and sum — but first check whether any source dominates the output node directly.

$V_s$ acting alone (current source open-circuited). $V_s$ is connected directly across the same two rails as $V_o$, with no series impedance between them, so $$V_o^{(V_s)}(t)=V_s(t)=10\sin(20t)\ \text{V}.$$
$i_s$ acting alone (voltage source killed — replaced by a short). Killing $V_s$ places an ideal short directly across the very same two rails that $V_o$ is measured on, forcing $$V_o^{(i_s)}(t)=0\ \text{V}$$ regardless of $i_s$, $R$, $L$, or $C$ — none of that current can develop a voltage across a short.
Superpose. $$V_o(t)=V_o^{(V_s)}(t)+V_o^{(i_s)}(t)=10\sin(20t)+0=\boxed{10\sin(20t)\ \text{V}}.$$
Cross-check. Direct nodal analysis of the full circuit confirms the same result: an ideal voltage source in parallel with a node is a hard constraint that fixes that node’s voltage exactly, independent of whatever current sources or impedances also terminate there — $R$, $L$, and $i_s$ only affect how much current $V_s$ itself must supply, never $V_o$.

Quantity	Result
$V_o(t)$	$\boxed{10\sin(20t)\ \text{V}}$

Concept & Theory

Topic. Superposition with an ideal voltage source directly in parallel with the observed node.

Key relations. Superposition: response $=\sum$ individual-source responses with all other independent sources killed (voltage → short, current → open); an ideal parallel voltage source fixes its node voltage regardless of any other parallel element.

Why this works. Superposition is valid because the network is linear, but its power here is diagnostic: recognising that $V_s$ sits directly across the output with zero intervening impedance shows immediately that $V_s$ alone reproduces the full output, and that killing $V_s$ (a short across the same two rails) forces every other source’s contribution to exactly zero at that node.

Common pitfall. Grinding through full phasor nodal analysis for $R$, $L$, $C$, and $i_s$ without first noticing the direct short this produces at the output node once $V_s$ is killed — correct but a large waste of the allotted time.

Source. Sadiku, Fundamentals of Electric Circuits §4.3 (superposition), §9.6 (phasor networks).

Question 5 of 11: Superposition in an AC parallel network

Part A — Circuits

Question 5: Superposition in an AC parallel network [20]