Question 6 of 6: Laplace (s-domain) analysis of an RLC circuit
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.
Question 6: Laplace (s-domain) analysis of an RLC circuit [5 + 5 + 10]
Figure‑6 — Source‑switched RLC circuit with stored energy i_L(0⁻)=2 A and V_c(0⁻)=6 V.
Given. A $15\text{ V}$ step drives $R=4\,\Omega$ in series with a parallel $L\!\parallel\!C$; $L=5\text{ H}$ (with $i_L(0^{-})=2\text{ A}$) and $C=0.1\text{ F}$ (with $V_c(0^{-})=6\text{ V}$). $V_c$ is the voltage across the parallel section.
Quantity
Value
Quantity
Value
Step input
$15\text{ V}\Rightarrow15/s$
$R$
$4\,\Omega$
$L$
$5\text{ H}$
$i_L(0^{-})$
$2\text{ A}$
$C$
$0.1\text{ F}$
$V_c(0^{-})$
$6\text{ V}$
Find. (a) the s-domain model; (b) $V_c(s)$; (c) $V_c(t)$.
Approach. Replace each element by its s-domain admittance with an initial-condition source, write one nodal equation at the $V_c$ node, then invert with the damped-sinusoid transform pair.
Laplace equivalent (part a). The step source becomes $15/s$ behind $R=4$. The inductor becomes admittance $\dfrac{1}{sL}=\dfrac{1}{5s}$ in parallel with a current source $\dfrac{i_L(0^{-})}{s}=\dfrac{2}{s}$; the capacitor becomes admittance $sC=0.1s$ in parallel with a current source $C\,V_c(0^{-})=0.6\text{ A}$.
Nodal equation at the $V_c$ node (part b). Summing currents leaving the node: $$\frac{V_c-15/s}{4}+\frac{V_c}{5s}+\frac{2}{s}+0.1s\,V_c-0.6=0.$$
Solve for $V_c(s)$ (part b). Collecting $V_c$: $\left(0.1s+0.25+\dfrac{0.2}{s}\right)V_c=\dfrac{15}{4s}-\dfrac{2}{s}+0.6.$ Multiplying through by $s$ and simplifying (the constant $\tfrac{15}{4}-2=\tfrac74$): $$\boxed{V_c(s)=\frac{0.6s+1.75}{0.1s^{2}+0.25s+0.2}=\frac{6s+17.5}{s^{2}+2.5s+2}.}$$
Pole locations (part c). $s^{2}+2.5s+2=0\Rightarrow s=-1.25\pm j0.6614$, an under-damped pair ($\alpha=1.25$, $\omega_d=0.6614\text{ rad/s}$). There is no pole at $s=0$, so $V_c(\infty)=0$ — the inductor shorts the node at DC, as expected.
Invert to the time domain (part c). Write $V_c(t)=e^{-1.25t}\big[A\cos(0.6614t)+B\sin(0.6614t)\big]$. Continuity gives $A=V_c(0^{+})=6$. The initial slope follows from $i_C(0^{+})=i_R-i_L=\dfrac{15-6}{4}-2=0.25\text{ A}$, so $\dfrac{dv_c}{dt}(0^{+})=\dfrac{i_C}{C}=2.5\text{ V/s}=-\alpha A+\omega_d B$, giving $B=\dfrac{2.5+1.25\cdot6}{0.6614}=15.12$. Hence $$\boxed{V_c(t)=e^{-1.25t}\big[6\cos(0.6614t)+15.12\sin(0.6614t)\big]\text{ V},\quad t\ge0.}$$
