Question 3 of 6: First-order RC switching transient
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.
Figure 3 — $V_s=15\text{ V}$ behind a $5\,\Omega$ feeds a horizontal $5\,\Omega$ to the load node (a $5\,\Omega$ in parallel with $C=0.5\text{ F}$). In position a the switch is open across the horizontal $5\,\Omega$ (current flows through it); in position b it shorts that $5\,\Omega$, connecting the source node directly to the load.
Given. $V_s=15\text{ V}$; source resistor $5\,\Omega$ (always in circuit); a horizontal $5\,\Omega$ between the source node and the load node; load = $5\,\Omega\,\|\,C$ with $C=0.5\text{ F}$. Position a (for $t<0$) routes current through the horizontal $5\,\Omega$; position b (for $t\ge0$) shorts it, tying the source node to the load node.
Approach. A capacitor voltage is continuous, so find the pre-switch steady state (cap open) for the initial value, the post-switch steady state for the final value, and the post-switch Thévenin resistance for the time constant.
Initial value (pos a, $t<0$). At DC steady state the capacitor is an open circuit, so the current runs $15\text{ V}\to5\,\Omega\text{(source)}\to 5\,\Omega\text{(horiz.)}\to5\,\Omega\text{(load)}$. The load node is a simple divider of three equal resistors: $$v_c(0^-)=15\cdot\frac{5}{5+5+5}=5\text{ V}.$$ By continuity $\boxed{v_c(0^+)=5\text{ V}}$.
Final value (pos b, $t\to\infty$). The horizontal $5\,\Omega$ is shorted, so the source node is the load node. With the cap open, the load node is a two-resistor divider: $$v_c(\infty)=15\cdot\frac{5}{5+5}=\boxed{7.5\text{ V}}.$$
Time constant. Kill $V_s$ (short) and look back from the capacitor in position b: the source $5\,\Omega$ and the load $5\,\Omega$ appear in parallel, $$R_{th}=5\,\|\,5=2.5\,\Omega,\qquad \tau=R_{th}C=2.5\times0.5=1.25\text{ s}.$$
Initial slope and capacitor current. For a single-time-constant response, $\dot v_c(0^+)=\dfrac{v_c(\infty)-v_c(0^+)}{\tau}=\dfrac{7.5-5}{1.25}=\boxed{2\text{ V/s}}$, so $i_c(0^+)=C\,\dot v_c(0^+)=0.5\times2=\boxed{1\text{ A}}$. (Check by KCL at $t=0^+$: in-current $(15-5)/5=2\text{ A}$, load-current $5/5=1\text{ A}$, difference $=1\text{ A}$ into the cap. ✓)
