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, mesh/nodal analysis with dependent sources, first-order RC transients, AC phasor analysis, Thévenin equivalents, maximum-power transfer and Laplace (s-domain) analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of super-node bookkeeping and 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; a sine (peak) time reference is used in Q4 to match the source $100\sin(2\pi 60\,t)$, and the sources marked rms in Q5 are treated as rms.
Question 4: AC steady-state phasor analysis [10 + 10]
Figure 4 — AC circuit: two identical $R+j\omega L$ branches in parallel, then a series capacitor carrying $v_c(t)$.
Given. Source $v_s(t)=100\sin(2\pi 60\,t)\text{ V}$; a $10\,\Omega$ series resistor; two identical parallel branches each $R+j\omega L$; then a series capacitor $C=10\,\mu\text{F}$ across which $v_c$ is taken.
Quantity
Value
Quantity
Value
$V_m$
$100\text{ V (peak)}$
$R_1=R_2$
$1\,\Omega$
$f$
$60\text{ Hz}$
$L_1=L_2$
$4\text{ mH}$
$\omega=2\pi f$
$376.99\text{ rad/s}$
$C$
$10\,\mu\text{F}$
Find. (a) $i_s(t)$ and $v_c(t)$; (b) the phasor diagram.
Approach. Convert every element to an impedance, combine the identical parallel branches, add the series $10\,\Omega$ and capacitor to get the driving-point impedance, then get $\overline{I}_s$ and $\overline{V}_c$ by Ohm’s law and a voltage split.
Branch impedances. $\omega L=376.99\cdot4\times10^{-3}=1.508\,\Omega$, so each parallel branch is $Z_{br}=1+j1.508\,\Omega$. The two identical branches in parallel halve this: $Z_p=\tfrac12 Z_{br}=0.500+j0.754\,\Omega$.
Capacitor and total impedance. $Z_C=\dfrac{1}{j\omega C}=\dfrac{1}{j(376.99)(10^{-5})}=-j265.26\,\Omega$. Adding the series pieces,
$$Z_{\text{tot}}=10+Z_p+Z_C=10.5-j264.5\,\Omega=264.7\angle{-87.73^\circ}\,\Omega.$$
Source current, part (a). With $\overline{V}_s=100\angle0^\circ\text{ V}$ (peak reference),
$$\overline{I}_s=\frac{\overline{V}_s}{Z_{\text{tot}}}=\frac{100\angle0^\circ}{264.7\angle{-87.73^\circ}}=0.378\angle{87.73^\circ}\text{ A}\;\Longrightarrow\;\boxed{i_s(t)=0.378\sin(2\pi 60\,t+87.7^\circ)\text{ A}.}$$
The near-$+90^\circ$ angle confirms the network is strongly capacitive (the $265\,\Omega$ capacitor dominates).
Capacitor voltage, part (a). $\overline{V}_c=\overline{I}_s Z_C=(0.378\angle87.73^\circ)(265.26\angle{-90^\circ})$:
$$\overline{V}_c=100.2\angle{-2.27^\circ}\text{ V}\;\Longrightarrow\;\boxed{v_c(t)=100.2\sin(2\pi 60\,t-2.27^\circ)\text{ V}.}$$
Almost all of the source voltage appears across the capacitor, as expected when $|Z_C|\gg$ the rest.
Phasor diagram, part (b). Take $\overline{V}_s$ along the reference axis. $\overline{I}_s$ leads it by $87.7^\circ$ (nearly quadrature, capacitive); $\overline{V}_c$ lags $\overline{I}_s$ by exactly $90^\circ$ and therefore sits just $2.3^\circ$ below the real axis, essentially collinear with $\overline{V}_s$.
Phasor diagram (part b) — $\overline{V}_s$ reference; $\overline{I}_s$ leads by 87.7° (capacitive); $\overline{V}_c$ trails $\overline{I}_s$ by 90°, nearly along $\overline{V}_s$. Current shown on its own scale.
