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 reduction, node/mesh 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; the cosine time reference is used throughout Q4 and Q5 (so $10\sin(377t+90^\circ)=10\cos 377t$ becomes $10\angle 0^\circ$), and the sources marked rms in Q5 are treated as rms.
Figure‑3 — First‑order RC circuit. The switch (parallel with R1) is closed for a long time, then opens at t = 0.
Given. A $15\text{ V}$ source feeds node $X$; between $X$ and node $Y$ sit $R_1=3\,\Omega$ and the switch in parallel. From $Y$: $R_2=5\,\Omega$ to ground and $R_3=2\,\Omega$ to the capacitor node, with $C=2\text{ mF}$ to ground. $V_c$ is the capacitor voltage.
Quantity
Value
Quantity
Value
Source
$15\text{ V}$
$R_1$ (∥ switch)
$3\,\Omega$
$R_2$
$5\,\Omega$
$R_3$
$2\,\Omega$
$C$
$2\text{ mF}$
Switch action
opens at $t=0$
Find. (a) $V_c(0^{+})$; (b) $\dfrac{dv_c}{dt}(0^{+})$; (c) $V_c(t)$; (d) the sketch.
Approach. Use $V_c$ continuity for $0^{+}$, a KCL at node $Y$ for the initial slope, and the final value plus Thevenin time constant for the full response.
Initial capacitor voltage (part a). For $t<0$ the closed switch shorts $R_1$, so $V_Y=V_X=15\text{ V}$; at DC steady state the capacitor is an open circuit, so no current flows in $R_3$ and there is no drop across it. Hence $V_c(0^{-})=V_Y=15\text{ V}$, and by continuity $$\boxed{V_c(0^{+})=15\text{ V}.}$$
Initial slope (part b). Just after the switch opens, $R_1$ is back in circuit and $V_c=15\text{ V}$ still. KCL at node $Y$ (with the cap node held at $15\text{ V}$): $\dfrac{15-V_Y}{3}=\dfrac{V_Y}{5}+\dfrac{V_Y-15}{2}\Rightarrow V_Y=12.10\text{ V}.$ The capacitor current is $i_C(0^{+})=\dfrac{V_Y-V_c}{R_3}=\dfrac{12.10-15}{2}=-1.452\text{ A}$, so $$\boxed{\frac{dv_c}{dt}(0^{+})=\frac{i_C(0^{+})}{C}=\frac{-1.452}{2\times10^{-3}}=-725.8\text{ V/s}.}$$
Final value. As $t\to\infty$ the capacitor is again open; with the switch open the current runs $15\to R_1\to R_2$, and $V_c(\infty)=V_Y=15\cdot\dfrac{R_2}{R_1+R_2}=15\cdot\dfrac{5}{8}=9.375\text{ V}.$
Time constant. Kill the source and look back from the capacitor: $R_{th}=R_3+(R_1\,\|\,R_2)=2+\dfrac{3\cdot5}{8}=3.875\,\Omega$, so $\tau=R_{th}C=3.875\cdot2\times10^{-3}=7.75\text{ ms}.$
Complete response (part c). With $V_c(0^{+})=15$ and $V_c(\infty)=9.375$: $$\boxed{V_c(t)=9.375+5.625\,e^{-t/7.75\text{ ms}}=9.375+5.625\,e^{-129.0\,t}\text{ V},\quad t\ge0.}$$ The initial slope $-\dfrac{5.625}{\tau}=-725.8\text{ V/s}$ matches part (b), confirming the fit.
Sketch (part d). A decaying exponential from $15\text{ V}$ to the $9.375\text{ V}$ asymptote, essentially settled after $5\tau\approx39\text{ ms}$:
Sketch of V_c(t): an exponential decay from 15 V (t = 0⁺) to the steady value 9.375 V, time constant τ = 7.75 ms.
