Question 3 of 6: First‑order RC transient after a switch opens
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination — 16‑Elec‑A1 Circuits, December 2019. Closed‑book, 3 hours; any five of the six questions constitute a complete paper (all six are solved here). A short Laplace‑transform table and star–delta / maximum‑power formulae are supplied with the exam.
Reference texts. C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (McGraw‑Hill); W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis, 9th ed. (McGraw‑Hill). Methods: series/parallel & Δ–Y reduction, nodal analysis, first‑order and second‑order transients, AC phasor / Thévenin analysis, maximum‑power transfer, superposition, and the Laplace‑transform method.
Reconstructed figures. Each network below is redrawn element‑by‑element from the paper’s figures; every reference node, source polarity, switch state and current direction used in the solution is stated alongside it so a reader can reproduce every sign. Where a component placement could not be read with certainty it is flagged in a Check note — check those against your own copy of the paper.
Question 3: First‑order RC transient after a switch opens [4+4+2+2 + 8]
[Figure not reproduced: Figure‑3 — redrawn single‑loop RC circuit. A 25 V source feeds a 5 Ω resistor, then a switch in parallel with 2 Ω, then a 3 Ω resistor in parallel with the 0.5 F capacitor ($v_c$ across it). The switch is closed for $t. See the official exam paper.]
Given. Series loop: 25 V source → 5 Ω → (switch ∥ 2 Ω) → (3 Ω ∥ $C$), with $C=0.5$ F and $v_c$ across the 3 Ω ∥ $C$ group. For $t<0$ the closed switch short‑circuits the 2 Ω.
Element
Value
Element
Value
Source
$25$ V dc
Series $R$
$5\,\Omega$
Switched $R$
$2\,\Omega$
Load $R$
$3\,\Omega$
Capacitor
$0.5$ F
Switch
closed $t<0$, opens $t=0$
Find. $v_c(0^+),\ \dot v_c(0^+),\ i_c(0^+),\ v_c(\infty)$ and $v_c(t),\ t\ge0$.
Approach. Use capacitor‑voltage continuity to carry $v_c$ across the switching instant, find the two steady states (cap open) before and after, get $\tau=R_{\text{th}}C$ from the post‑switch network, and assemble the single‑exponential response.
Initial condition ($t<0$, switch closed). The closed switch shorts the 2 Ω; in steady state the capacitor is open, so current flows $25\,\text{V}\to5\,\Omega\to3\,\Omega$: $I=\tfrac{25}{5+3}=3.125$ A and $$v_c(0^-)=I\cdot3=9.375\ \text{V}.$$ By continuity $\boxed{v_c(0^+)=9.375\ \text{V}.}$
Final value ($t\to\infty$, switch open). Now the 2 Ω is in the loop and the cap is again open: $I_\infty=\tfrac{25}{5+2+3}=2.5$ A, so $$\boxed{v_c(\infty)=I_\infty\cdot3=7.5\ \text{V}.}$$
Capacitor current at $0^+$. Just after opening, the group voltage is still $9.375$ V. The current delivered through $5+2=7\,\Omega$ is $I_{\text{tot}}=\tfrac{25-9.375}{7}=2.232$ A, while the 3 Ω draws $\tfrac{9.375}{3}=3.125$ A; the difference is the cap current: $$\boxed{i_c(0^+)=2.232-3.125=-0.893\ \text{A}.}$$
Time constant. Kill the source (short it); the cap sees $3\,\|\,(5+2)=3\,\|\,7=2.1\,\Omega$, so $$\tau=R_{\text{th}}C=2.1\times0.5=1.05\ \text{s}.$$
Assemble $v_c(t)$. With the first‑order form $v_c(t)=v_c(\infty)+[v_c(0^+)-v_c(\infty)]e^{-t/\tau}$: $$\boxed{v_c(t)=7.5+1.875\,e^{-t/1.05}\ \text{V},\quad t\ge0.}$$ Its initial slope $-1.875/1.05=-1.79$ V/s matches step 4 — a good consistency check.
