Superposition theorem: 23-Mechatronics-A2 Circuits and Electronics, December 2018, Q2

Question 2: Superposition theorem [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.

Element	Value
Left source	20 V (+ at top)
Right source	30 V (+ at top)
Current source (bridges node 1–node 2)	5 A, arrow into node 2 side
Series resistors	3 Ω (left), 5 Ω (right)
Middle resistor ($V_o$ across it)	4 Ω, node 1–node 2
Shunt resistors to ground	2 Ω (node 1), 4 Ω (node 2)

Find. $V_o$ (voltage across the middle 4 Ω resistor, − at node 1, + at node 2) using superposition.

Figure-2: 20 V and 30 V sources with series resistors (3 Ω, 5 Ω), a 5 A current source bridging the two interior nodes, and shunt resistors (2 Ω, 4 Ω) to the common bottom rail. $V_o$ is measured across the middle 4 Ω resistor, − at node 1 (left) and + at node 2 (right).

Approach. Superposition: solve $V_o$ with only the 20 V source active (30 V shorted, 5 A opened), then only the 30 V source active, then only the 5 A source active, and add the three contributions.

20 V source alone (30 V → short, 5 A → open). Nodal analysis at node 1, node 2 with the current source removed: $$\frac{20-V_1}{3}=\frac{V_1}{2}+\frac{V_1-V_2}{4},\qquad \frac{V_1-V_2}{4}+\frac{0-V_2}{5}=\frac{V_2}{4}.$$ Solving gives $V_1=6.06\ \text{V}$, $V_2=1.75\ \text{V}$, so $$V_o^{(20\text{V})}=V_2-V_1=\boxed{-4.31\ \text{V}}.$$
30 V source alone (20 V → short, 5 A → open). Same network with the right source active: solving the analogous pair gives $$V_o^{(30\text{V})}=\boxed{7.19\ \text{V}}.$$
5 A source alone (both voltage sources shorted). With 5 A injected from node 1 into node 2: $$V_o^{(5\text{A})}=\boxed{2.75\ \text{V}}.$$
Superpose. $$V_o=V_o^{(20\text{V})}+V_o^{(30\text{V})}+V_o^{(5\text{A})}=-4.31+7.19+2.75=\boxed{5.63\ \text{V}}.$$
Cross-check. Solving the full circuit directly (all three sources active at once, single nodal solve) gives $V_1=15.69\ \text{V}$, $V_2=21.32\ \text{V}$, $V_o=V_2-V_1=5.63\ \text{V}$ — matching the superposition sum.

Quantity	Result
Contribution of 20 V alone	$-4.31\ \text{V}$
Contribution of 30 V alone	$7.19\ \text{V}$
Contribution of 5 A alone	$2.75\ \text{V}$
Total	$\boxed{V_o=5.63\ \text{V}}$

Concept & Theory

Topic. Superposition theorem in a linear resistive network with mixed independent sources.

Key relations. Superposition: for a linear circuit, the response to several independent sources equals the sum of the responses to each acting alone (voltage sources shorted, current sources opened when inactive). Nodal KCL used per sub-circuit.

Why this works. Linearity (Ohm’s law and KCL/KVL are linear operators) means the network response is a linear combination of the individual source contributions; deactivating a source sets its contribution to zero without altering the topology seen by the others.

Common pitfall. Forgetting to re-solve the whole reduced network (not just a single branch) for each source, and sign errors when a shorted/opened source changes which resistors become parallel/series.

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

Question 2 of 11: Superposition theorem

Question 2: Superposition theorem [20]