Question 6 of 6: Second‑order transient by the Laplace transform
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 6: Second‑order transient by the Laplace transform [10 + 10]
[Figure not reproduced: Figure‑6 — redrawn transient circuit (time domain). A 15 V source, closed at $t=0$, feeds a 4 Ω series resistor into a node where $C=0.2$ F ($V_c$, with $V_c(0)=5$ V) and $L=1$ H ($i_L(0)=1$ A) are in parallel to ground. See the official exam paper.]
Given. For $t\ge0$: $15$ V dc through a series $4\,\Omega$ into node X, with $C=0.2$ F and $L=1$ H in parallel from X to ground. Initial conditions $V_c(0)=5$ V and $i_L(0)=1$ A (downward). $V_c$ is the node X voltage across the capacitor.
Element
Value
Initial condition
Value
Source
$15$ V dc
$V_c(0)$
$5$ V
Series $R$
$4\,\Omega$
$i_L(0)$
$1$ A
$C$
$0.2$ F
$L$
$1$ H
Find. The $s$‑domain circuit and $V_c(t)$ for $t\ge0$.
Approach. Transform each element to its impedance with an initial‑condition source, write one node equation for $V_c(s)$, and invert with the damped‑sinusoid Laplace pairs.
(a) Laplace‑transformed circuit. The step source becomes $15/s$; the resistor stays $4$; the capacitor becomes an impedance $1/(sC)=5/s$ in series with a voltage source $V_c(0)/s=5/s$; the inductor becomes an impedance $sL=s$ in series with a source $L\,i_L(0)=1$ V:
Laplace‑domain equivalent for $t\ge0$. Capacitor branch: $5/s$ impedance in series with the $V_c(0)/s=5/s$ initial‑condition source; inductor branch: $s$ impedance in series with the $L\,i_L(0)=1$ V source. The node voltage is $V_c(s)$.
(c) Solve for $V_c(s)$ and invert.
Node equation at X. Summing currents leaving (resistor, capacitor, inductor) to zero, with the IC sources built into the capacitor and inductor terms: $$\frac{V_c-15/s}{4}+0.2\big(sV_c-5\big)+\Big(\frac{V_c}{s}+\frac1s\Big)=0.$$
Collect $V_c(s)$. Grouping and multiplying through by $s$: $$V_c(s)\,\big(0.2s^2+0.25s+1\big)=s+2.75.$$
Closed form. Multiplying numerator and denominator by 5, $$\boxed{V_c(s)=\frac{5s+13.75}{s^2+1.25s+5}.}$$ The initial‑ and final‑value theorems confirm $V_c(0^+)=5$ V and $V_c(\infty)=0$ (the inductor shorts the node at DC).
Complete the square. $s^2+1.25s+5=(s+0.625)^2+(2.147)^2$, so $\alpha=0.625$, $\omega_d=\sqrt{5-0.625^2}=2.147$ rad/s (under‑damped). Writing the numerator about the shifted pole, $5s+13.75=5(s+0.625)+10.625$.
Invert. Using $e^{-\alpha t}\cos\omega_d t\leftrightarrow\frac{s+\alpha}{(s+\alpha)^2+\omega_d^2}$ and the matching sine pair, $$\boxed{V_c(t)=e^{-0.625t}\big[\,5\cos(2.147t)+4.95\sin(2.147t)\,\big]\ \text{V},\quad t\ge0.}$$ (The sine coefficient is $10.625/2.147=4.95$.)
