Question 4 of 7: Sinusoidal Steady State — Phasors and Capacitor Current
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam — 04-BS-4 Electric Circuits and Power — undated sitting (internal
evidence places it at May 2019). 3 hours, closed book, one
double-sided aid sheet, approved Casio/Sharp calculator only. The exam instructs
"any five questions constitute a complete paper" — all seven are solved below as a
complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits
(7th ed.) — Ch. 2–4 (resistive circuits, node/mesh analysis, Thévenin/Norton,
maximum power transfer), Ch. 7 (first-order RC/RL transients), Ch. 9–10 (sinusoidal
steady state, phasors, AC power); Chapman, Electric Machinery Fundamentals
(5th ed.) — Ch. 1 (magnetic circuits, reluctance, fringing, force on an armature);
Sadiku, Elements of Electromagnetics — Ch. 5, 8 (magnetic energy and force);
Mano & Ciletti, Digital Design — Ch. 2–3 (Boolean algebra, combinational
logic design).
Question 4: Sinusoidal Steady State — Phasors and Capacitor Current (20 marks)
Given. Three branches — $R_1$ in series with $L_1$, $R_2$ alone, and $C$ alone — all in parallel, driven by the current source $i_s(t)=400\cos(20000t)\text{ mA}$.
Given data
Quantity
Value
$R_1$
$600\ \Omega$
$L_1$
$60\text{ mH}$
$R_2$
$200\ \Omega$
$C$
$125\text{ nF}$
$\omega$
$20000\text{ rad/s}$
$i_s(t)$
$400\cos(20000t)\text{ mA}$
Find. The phasor-domain circuit, the impedance $Z_{eq}$ seen by the source, the capacitor-current phasor $\underline{I_c}$, and $i_c(t)$.
Figure 4 — AC steady-state circuit for Question 4.
Approach. Convert every element to its impedance at $\omega=20000\text{ rad/s}$, combine the three parallel branches into one equivalent impedance, use $\underline{I_s}\,Z_{eq}$ to get the (single) node-voltage phasor, then divide by $Z_C$ to get $\underline{I_c}$ and convert back to the time domain.
Part (a) — phasor-domain impedances. Using $\underline{I_s}=0.4\angle0^\circ\text{ A}$ (amplitude form, matching the given cosine) and $\omega=20000\text{ rad/s}$: $$Z_{R_1L_1}=R_1+j\omega L_1=600+j(20000)(0.06)=600+j1200\ \Omega=1341.6\angle63.43^\circ\ \Omega$$$$Z_{R_2}=200\ \Omega\qquad\qquad Z_C=\dfrac{1}{j\omega C}=\dfrac{1}{j(20000)(125\times10^{-9})}=-j400\ \Omega=400\angle-90^\circ\ \Omega$$ The phasor circuit is these three impedances in parallel, driven by $\underline{I_s}$.
Part (b) — equivalent impedance seen by the source. All three branches sit between the same two nodes, so they combine as admittances in parallel: $$Y_{eq}=\dfrac{1}{Z_{R_1L_1}}+\dfrac{1}{Z_{R_2}}+\dfrac{1}{Z_C}=\dfrac{1}{600+j1200}+\dfrac{1}{200}+\dfrac{1}{-j400}$$$$Z_{eq}=\dfrac{1}{Y_{eq}}=(167.69-j57.64)\ \Omega=\boxed{177.3\angle-18.97^\circ\ \Omega}$$
Step — node-voltage phasor. All three branches share the same terminal voltage $\underline{V}$, driven directly by the current source through $Z_{eq}$: $$\underline{V}=\underline{I_s}\,Z_{eq}=(0.4\angle0^\circ)(177.3\angle-18.97^\circ)=70.93\angle-18.97^\circ\text{ V}$$
Part (c) — capacitor current phasor. $$\underline{I_c}=\dfrac{\underline{V}}{Z_C}=\dfrac{70.93\angle-18.97^\circ}{400\angle-90^\circ}=\boxed{0.1773\angle71.03^\circ\text{ A}}=177.3\angle71.03^\circ\text{ mA}$$ As a check, the branch currents through $R_1$-$L_1$ ($52.87\angle{-}82.41^\circ\text{ mA}$) and $R_2$ ($354.6\angle{-}18.97^\circ\text{ mA}$) sum with $\underline{I_c}$ to $400.0\angle0.0^\circ\text{ mA}=\underline{I_s}$, confirming KCL at the node.
Part (d) — capacitor current in the time domain. Since a positive-frequency phasor $\underline{I_c}=|I_c|\angle\theta$ corresponds to $i_c(t)=|I_c|\cos(\omega t+\theta)$: $$\boxed{i_c(t)=177.3\cos(20000t+71.03^\circ)\text{ mA}}$$
