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22-Elec-A1 Circuits · December 2018

Question 2 of 6: Superposition

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — series/parallel and Δ–Y reduction, superposition, node analysis with dependent sources, first-order RC transients, AC phasor/node analysis, Thévenin equivalents and maximum-power transfer, and Laplace (s-domain) analysis of second-order circuits; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the initial-condition source models used in Q6.

Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here as a study resource. A Laplace-transform table and a star–delta conversion table are supplied on the last two pages of the exam. Phasor magnitudes are written as magnitude $\angle$ angle with angles in degrees; because $e$ and $i$ in Q5 are written as $40\sqrt2\cos$ and $5\sqrt2\cos$ (peak $=\sqrt2\times$rms), rms phasors $\mathbf E=40\angle30^\circ$ V and $\mathbf I=5\angle0^\circ$ A are used so the computed power is in watts directly.

Question 2: 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.

+−20 V3ΩN12Ω4Ω− V₀ +5 AN24Ω5ΩN3+−30 V
Figure‑2 — three independent sources; V₀ is measured across the middle 4Ω (− at N1, + at N2). Bottom rail is the reference.

Given. Three independent sources act on the network: a $20\text{ V}$ source in series with $3\,\Omega$ feeding node N1, a $30\text{ V}$ source in series with $5\,\Omega$ feeding node N2 (through node N3), a $5\text{ A}$ source directed from N1 to N2, and two grounded resistors ($2\,\Omega$ at N1, $4\,\Omega$ at N2). $V_o$ is the voltage across the middle $4\,\Omega$ with the − terminal at N1 and $+$ at N2, i.e. $V_o=V_{N2}-V_{N1}$.

SourceValueResistorValue
Left voltage source$20\text{ V}$Series (left)$3\,\Omega$
Right voltage source$30\text{ V}$Series (right)$5\,\Omega$
Current source (N1$\to$N2)$5\text{ A}$N1–gnd / N2–gnd$2\,\Omega$ / $4\,\Omega$
Middle resistor ($V_o$)$4\,\Omega$

Find. $V_o$ by superposition (contribution of each source, then sum).

Approach. Deactivate two sources at a time (voltage sources shorted, current source opened), solve the two-node network for each acting source in turn, then add the three partial results.

  1. Node framework. With the bottom rail as reference and unknowns $V_1$ (N1) and $V_2$ (N2), general KCL gives $$\Big(\tfrac13+\tfrac12+\tfrac14\Big)V_1-\tfrac14 V_2=\tfrac{20}{3}-5,\qquad -\tfrac14 V_1+\Big(\tfrac14+\tfrac14+\tfrac15\Big)V_2=\tfrac{30}{5}+5,$$ where the right-hand terms carry each source. Superposition solves this by switching the terms on and off.
  2. Only the $20\text{ V}$ source active (short $30\text{ V}$, open $5\text{ A}$). The reduced system gives $V_1'=6.707\text{ V}$, $V_2'=2.395\text{ V}$, so $$V_o'=V_2'-V_1'=-4.31\text{ V}.$$
  3. Only the $30\text{ V}$ source active (short $20\text{ V}$, open $5\text{ A}$). This gives $V_1''=2.156\text{ V}$, $V_2''=9.341\text{ V}$, so $$V_o''=9.341-2.156=+7.19\text{ V}.$$
  4. Only the $5\text{ A}$ source active (short both voltage sources). This gives $V_1'''=-3.234\text{ V}$, $V_2'''=5.988\text{ V}$, so $$V_o'''=5.988-(-3.234)=+9.22\text{ V}.$$
  5. Superpose. Add the three contributions: $$\boxed{V_o=-4.31+7.19+9.22=12.10\text{ V}.}$$ A direct two-node solve of the full network gives $V_1=5.63\text{ V}$, $V_2=17.72\text{ V}$, $V_o=12.10\text{ V}$ — confirming the superposition sum.
Contribution$V_o$ part
From $20\text{ V}$ source$-4.31\text{ V}$
From $30\text{ V}$ source$+7.19\text{ V}$
From $5\text{ A}$ source$+9.22\text{ V}$
Total $V_o$$\mathbf{12.10\text{ V}}$