Question 4 of 6: Superposition with two different-frequency sources
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Elec-A1 Circuits, December 2014. Closed-book, 3-hour paper; a Laplace-transform table and Y–Δ formulas are supplied in the appendix. Six questions of equal value; any five constitute a complete paper. Full worked solutions to all six questions are given below so the set can be used for study regardless of which five a candidate chose.
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — series/parallel reduction and the Wheatstone bridge, mesh/supermesh and nodal analysis, first-order transients, AC steady-state, Thévenin’s theorem and maximum-power transfer, and Laplace-domain circuit analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the s-domain and initial-condition source models.
each network below is redrawn element-by-element from the original drawing. The Wheatstone bridge of Q1 is shown in the equivalent rectangular form (electrically identical to the diamond on the paper). Every reference node, mesh direction and source polarity used is stated with its solution so a reader can reproduce every sign. Phasor answers are quoted in magnitude∡angle form with angles in degrees; the source of Q5 is rms.
Question 4: Superposition with two different-frequency sources [20]
Given. An ideal voltage source $v_s=20\sin 5t\text{ V}$ and an ideal current source $i_s=10\cos(10t+30^{\circ})\text{ A}$ act at two different frequencies ($\omega=5$ and $\omega=10\text{ rad/s}$). The output $v_o$ is taken across the $10\,\Omega$ resistor, which is connected directly across $v_s$. The current-source branch adds $20\,\Omega$, $2\text{ H}$ and $1\text{ F}$.
Find. $v_o(t)$.
[Figure not reproduced: Figure-4 (redrawn). $v_s$, the $10\,\Omega$ (across which $v_o$ is measured) share the same node pair, so $v_s$ sits directly across the output. The two sources run at different frequencies, so superposition must treat them one at a time. See the official exam paper.]
Approach. Because the two sources are at different frequencies they must be superposed (phasors of different $\omega$ cannot be added). Kill one source at a time, find each contribution to $v_o$, then sum the time-domain results.
Contribution of $v_s$ (kill $i_s$: open it). With the current source opened, its branch ($20\,\Omega+2\text{ H}+1\text{ F}$) is broken and carries no current, so it cannot affect the left node. The ideal source $v_s$ is connected straight across the $10\,\Omega$, therefore $$v_{o}^{(1)}=v_s=20\sin 5t\ \text{V}.$$
Contribution of $i_s$ (kill $v_s$: short it). Replacing $v_s$ by a short circuit ties the top output node to ground through zero ohms, placing that short directly across the $10\,\Omega$. All of the current-source current returns through the $0\,\Omega$ short rather than the resistor, so the voltage across the $10\,\Omega$ is $$v_{o}^{(2)}=0\ \text{V}.$$
Superpose. Adding the two contributions, $$v_o(t)=v_{o}^{(1)}+v_{o}^{(2)}=\boxed{20\sin 5t\ \text{V}}.$$
Why the elaborate right-hand branch drops out. An ideal voltage source in parallel with the output clamps that node no matter what current the rest of the circuit injects. The $20\,\Omega$–$2\text{ H}$–$1\text{ F}$ branch and $i_s$ do carry current, but that current simply circulates through $v_s$; it produces zero volts across the $10\,\Omega$. The problem is a deliberate test of that principle rather than a numerically heavy phasor exercise.
