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22-Elec-A1 Circuits · Undated paper

Question 5 of 6: AC node-voltage analysis

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

Notes on this paper

National Exam — 16-Elec-A1 Circuits. 3 hours, closed-book; one approved Casio/Sharp calculator permitted. Any five of the six questions constitute a complete paper and all are of equal value — all six are solved in full below. A Laplace-transform table and a Δ–Y conversion set are supplied on the exam’s last two pages.

Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — series/parallel reduction, the Wheatstone bridge, nodal analysis, first-order RC/RL transients, phasor (AC) analysis, Thévenin’s theorem, maximum power transfer, and the Laplace-transform method; W. H. Hayt, Engineering Circuit Analysis (9th ed.) for the dependent-source and second-order (RLC) material.

Each reference node, source polarity and current direction is stated alongside the working so a reader can reproduce every sign.

Question 5: AC node-voltage analysis [9 + 9 + 2]

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.

V1V2V31H0.2F5Ωio(t)2Hvs(t)+−0.1Fis(t)
Figure 5 — $\omega=5\text{ rad/s}$. $v_s$ fixes node $V_1$; top branch $V_1$–$V_3$ is a series $1\text{ H}+0.2\text{ F}$; $5\,\Omega$ (carrying $i_o$) across $V_1$–$V_2$; $2\text{ H}$ across $V_2$–$V_3$; $0.1\text{ F}$ from $V_2$ to ground; $i_s$ injected into $V_3$.

Given. $\omega=5$. Using a cosine reference, $\sin5t=\cos(5t-90^\circ)$, so $\mathbf V_1=\mathbf V_s=10\angle{-90^\circ}\text{ V}$ (the source ties node $V_1$ directly to ground) and $\mathbf I_s=5\angle30^\circ\text{ A}$ into $V_3$. Branch impedances at $\omega=5$:

ElementBetweenImpedanceAdmittance
$1\text{ H}+0.2\text{ F}$ (series)$V_1$–$V_3$$j5+\frac{1}{j1}=j4\,\Omega$$-j0.25\text{ S}$
$5\,\Omega$$V_1$–$V_2$$5\,\Omega$$0.2\text{ S}$
$2\text{ H}$$V_2$–$V_3$$j10\,\Omega$$-j0.1\text{ S}$
$0.1\text{ F}$$V_2$–gnd$-j2\,\Omega$$j0.5\text{ S}$

Find. $v_1(t),v_2(t),v_3(t),i_o(t)$.

Approach. With $V_1$ known from the source, write phasor KCL at the two unknown nodes $V_2$ and $V_3$, solve, then read $i_o$ across the $5\,\Omega$.

  1. Known node. The ideal source sets $\mathbf V_1=10\angle{-90^\circ}=-j10\text{ V}$.
  2. KCL at $V_2$. $$\frac{V_2-V_1}{5}+\frac{V_2-V_3}{j10}+\frac{V_2}{-j2}=0.$$
  3. KCL at $V_3$ (source current enters): $$\frac{V_3-V_1}{j4}+\frac{V_3-V_2}{j10}=\mathbf I_s=5\angle30^\circ.$$
  4. Solve the $2\times2$ phasor system. $\mathbf V_2=2.94\angle{-132.9^\circ}\text{ V}$ and $\mathbf V_3=8.99\angle149.1^\circ\text{ V}$.
  5. Current $i_o$. Across the $5\,\Omega$ from $V_1$ to $V_2$: $$\mathbf I_o=\frac{V_1-V_2}{5}=\boxed{1.62\angle{-75.7^\circ}\text{ A}.}$$
  6. Back to the time domain (cosine reference, $\omega=5$):$$v_1(t)=10\sin5t\text{ V},\quad v_2(t)=2.94\cos(5t-132.9^\circ)\text{ V},$$$$v_3(t)=8.99\cos(5t+149.1^\circ)\text{ V},\quad i_o(t)=1.62\cos(5t-75.7^\circ)\text{ A}.$$
QuantityPhasorTime domain
$V_1$$10\angle{-90^\circ}$$10\sin5t\text{ V}$
$V_2$$2.94\angle{-132.9^\circ}$$2.94\cos(5t-132.9^\circ)\text{ V}$
$V_3$$8.99\angle149.1^\circ$$8.99\cos(5t+149.1^\circ)\text{ V}$
$i_o$$1.62\angle{-75.7^\circ}$$1.62\cos(5t-75.7^\circ)\text{ A}$