Capacitor voltage by superposition: 22-Elec-A1 Circuits, December 2019, Q5

Question 5: Capacitor voltage by superposition [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.

[Figure not reproduced: Figure‑5 — redrawn parallel AC network. Five branches share one node pair: a current source $i_s=5\cos(20t+10^\circ)$ A, a $10\,\Omega$ resistor, a $0.5$ F capacitor (output $V_o$), a $2$ H inductor, and an ideal voltage source $V_s=10\sin20t$ V. All operate at $\omeg. See the official exam paper.]

Given. All five elements are connected in parallel between the same two nodes; $V_o$ is the voltage across the $0.5$ F capacitor. Both sources act at $\omega=20$ rad/s: $i_s=5\cos(20t+10^\circ)$ A and the ideal $V_s=10\sin20t$ V.

Find. $V_o(t)$, obtained as the sum of the responses to $i_s$ acting alone and $V_s$ acting alone.

Approach. Apply superposition: deactivate one source at a time. The decisive observation is that an ideal voltage source sits directly across the output node, so it clamps $V_o$ regardless of the other elements.

Response to $i_s$ alone (kill $V_s$). Deactivating an ideal voltage source replaces it with a short‑circuit. That short is placed directly across the output node, so it also shorts the capacitor: the entire current $i_s$ returns through the short and $$V_{o,1}(t)=0.$$ None of the R, L, C values enter — the short wins.
Response to $V_s$ alone (kill $i_s$). Deactivating the current source opens its branch. The ideal $V_s$ is still connected directly across the capacitor, so it fixes the node voltage exactly: $$V_{o,2}(t)=V_s=10\sin20t\ \text{V}.$$
Superpose. $$\boxed{V_o(t)=V_{o,1}+V_{o,2}=10\sin20t\ \text{V}.}$$ The current source, resistor, inductor and capacitor all draw current from $V_s$ but cannot alter the clamped node voltage; their only effect is on the current the ideal source must supply.

Check — ideal‑source clamp. This reading takes the $V_s$ branch as an ideal source directly in parallel with the output (as the reconstruction shows), which makes $V_o=V_s$ exactly and the current‑source contribution zero — a recurring “trick” in this exam series. If your copy places a series element (e.g. the 2 H inductor) between $V_s$ and the capacitor node, redo step 2 as a phasor divider at $\omega=20$ with $Z_C=-j0.1\,\Omega$, $Z_L=j40\,\Omega$, $R=10\,\Omega$.

Contribution	Value at $\omega=20$
From $i_s$ alone	$0$
From $V_s$ alone	$10\sin20t\,\text{V}$
Total $V_o(t)$	$10\sin20t\,\text{V}$

Concept & Theory

Topic: superposition and the clamping action of an ideal voltage source.

Key relations: superposition (linear circuits): total response $=\sum$ single‑source responses; source deactivation — ideal $V$‑source → short, ideal $I$‑source → open.

Why this works: superposition holds because the network is linear. An ideal voltage source in parallel with a subnetwork fixes that node’s voltage independent of everything beside it, so the whole point of the problem is to recognise the clamp rather than grind phasor algebra.

Common pitfall: forgetting that shorting the killed voltage source also shorts anything in parallel with it (here the output), which is why $i_s$ contributes exactly zero — not a small residual.

Source: Sadiku, Fundamentals of Electric Circuits, 7th ed., §10.3 (superposition) & §4.3.

Question 5 of 6: Capacitor voltage by superposition

Question 5: Capacitor voltage by superposition [20]