Question 6 of 6: Laplace (s-domain) solution of a series RLC
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 and Δ–Y reduction, mesh/nodal 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; a sine (peak) time reference is used in Q4 to match the source $100\sin(2\pi 60\,t)$, and the sources marked rms in Q5 are treated as rms.
Question 6: Laplace (s-domain) solution of a series RLC [8 + 12]
Figure 6 — series RLC energised at $t=0$ with $i_L(0^{-})=0$, $v_c(0^{-})=5$ V.
Given. Series loop: $15\text{ V}$ dc source, switch (closes at $t=0$), $R=2\,\Omega$, $L=1\text{ H}$, $C=0.1\text{ F}$; initial conditions $i_L(0^{-})=0$, $v_c(0^{-})=5\text{ V}$ ($+$ on the top plate).
Find. (a) the s-domain circuit; (b) $V_c(t)$ for $t\ge0$.
Approach. Replace each element by its s-domain model (with initial-condition sources), write one mesh equation for $I(s)$, then form $V_c(s)$ and invert with the supplied transform table.
Laplace equivalent, part (a). The step source becomes $15/s$; the resistor stays $R=2$; the inductor becomes $sL=s$ with no series source because $i_L(0^{-})=0$; the capacitor becomes an impedance $1/(sC)=10/s$ in series with an initial-condition source $v_c(0^{-})/s=5/s$ (polarity opposing the charging current). The single-mesh s-domain circuit is:
s-domain equivalent — $15/s$ step source, $R=2$, $sL=s$, capacitor impedance $10/s$ in series with the $5/s$ initial-condition source.
Mesh equation. KVL around the loop for the mesh current $I(s)$, moving the initial-condition source to the left:
$$\frac{15}{s}-\frac{5}{s}=I(s)\left(R+sL+\frac{1}{sC}\right)=I(s)\left(2+s+\frac{10}{s}\right).$$
Hence $\dfrac{10}{s}=I(s)\,\dfrac{s^{2}+2s+10}{s}$, giving
$$I(s)=\frac{10}{s^{2}+2s+10}.$$
Capacitor voltage in the s-domain. $V_c(s)=\dfrac{1}{sC}I(s)+\dfrac{v_c(0^{-})}{s}=\dfrac{10}{s}\cdot\dfrac{10}{s^{2}+2s+10}+\dfrac{5}{s}=\dfrac{100}{s\,(s^{2}+2s+10)}+\dfrac{5}{s}.$
Partial fractions. Writing $\dfrac{100}{s(s^{2}+2s+10)}=\dfrac{A}{s}+\dfrac{Bs+C}{(s+1)^{2}+9}$ gives $A=10,\ B=-10,\ C=-20$. Combining with the $5/s$ term:
$$V_c(s)=\frac{15}{s}-\frac{10(s+1)+10}{(s+1)^{2}+3^{2}}.$$
Invert, part (b). Using $e^{-\alpha t}\cos\omega t\leftrightarrow\frac{s+\alpha}{(s+\alpha)^2+\omega^2}$ and $e^{-\alpha t}\sin\omega t\leftrightarrow\frac{\omega}{(s+\alpha)^2+\omega^2}$ with $\alpha=1,\ \omega=3$:
$$\boxed{V_c(t)=15-10\,e^{-t}\cos 3t-\tfrac{10}{3}\,e^{-t}\sin 3t\ \text{ V},\qquad t\ge0.}$$
The associated loop current is $i(t)=\tfrac{10}{3}e^{-t}\sin 3t\text{ A}$.
Checks. At $t=0$: $V_c=15-10=5\text{ V}$ (matches $v_c(0^{-})$, continuity). As $t\to\infty$: $V_c\to15\text{ V}$ (capacitor charges to the source). The complex roots $s=-1\pm j3$ mark an under-damped response ($\zeta=R/2\sqrt{C/L}=0.316$).
