Laplace-domain RLC response: 22-Elec-A1 Circuits, May 2013, Q6

Question 6: Laplace-domain RLC response [10 + 10]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $V_{in}=15\ \text{V}$ (dc), $R=3\ \Omega$ in series feeding a parallel $L=5\ \text{H}$ and $C=0.1\ \text{F}$; $i_L(0)=-1.5\ \text{A}$ (reference arrow upward, i.e. $+1.5\ \text{A}$ downward from the node), $v_c(0)=4\ \text{V}$.

Find. The s-domain circuit and $v_c(t)$ for $t\ge 0$.

Figure 7 — RLC network for Q6: 15 V source and 3 Ω feeding a parallel 5 H inductor and 0.1 F capacitor, both with the stated initial conditions.

Approach. Replace each element by its s-domain model (source $15/s$, resistor $R$, capacitor admittance $sC$ with an initial-condition current $Cv_c(0)$, inductor admittance $1/sL$ with an initial-condition current $i_L(0)/s$), write one node equation for $V_c(s)$, and invert by partial fractions.

(a) Transformed circuit. The source becomes $15/s$ behind $R=3$. The capacitor branch is an admittance $sC=0.1s$ in parallel with a current source $C\,v_c(0)=0.4$ (modelling the 4 V charge). The inductor branch is an admittance $1/(sL)=1/(5s)$ in parallel with a current source $i_L(0)/s=1.5/s$ (the initial current, positive downward).
Node equation at $V_c(s)$. KCL (current in from the source = currents into $C$ and $L$): $$\frac{15/s-V_c}{3}=\bigl(0.1sV_c-0.4\bigr)+\Bigl(\frac{V_c}{5s}+\frac{1.5}{s}\Bigr).$$
Solve for $V_c(s)$. Collecting terms and multiplying through by $s$: $$V_c(s)=\frac{0.4s+3.5}{0.1s^{2}+\tfrac{1}{3}s+0.2}=\frac{4s+35}{s^{2}+\tfrac{10}{3}s+2}.$$
Poles. $s^2+3.333s+2=0\Rightarrow s_{1}=-0.785,\ s_{2}=-2.549$ (both real, over-damped).
(b) Partial fractions and inversion for $v_c(t)$. Writing $V_c(s)=\dfrac{A}{s-s_1}+\dfrac{B}{s-s_2}$ gives $A=18.06$, $B=-14.06$ (note $A+B=4=v_c(0)$, as required). Therefore $$\boxed{v_c(t)=18.06\,e^{-0.785\,t}-14.06\,e^{-2.549\,t}\ \text{V},\quad t\ge 0.}$$ It starts at 4 V and decays to $0$ as the inductor becomes a dc short.

Quantity	Result
$V_c(s)$	$\dfrac{4s+35}{s^{2}+3.333s+2}$
Poles	$s=-0.785,\ -2.549\ \text{s}^{-1}$
Initial / final value	$v_c(0)=4\ \text{V}$, $v_c(\infty)=0$
$v_c(t)$	$\boxed{18.06e^{-0.785t}-14.06e^{-2.549t}\ \text{V}}$

Concept & Theory

Topic. Laplace-transform (s-domain) analysis of a second-order RLC circuit with initial conditions.

Key relations. Element models $Z_L=sL$ (or admittance $1/sL$ with source $i_L(0)/s$), $Z_C=1/sC$ (or admittance $sC$ with source $Cv_c(0)$); node analysis in $s$; inverse transform $\dfrac{A}{s-p}\leftrightarrow Ae^{pt}$.

Why this works. The Laplace transform turns the integro-differential node equation into an algebraic one, with the initial inductor current and capacitor voltage entering as independent sources. Real, distinct poles give an over-damped sum of two decaying exponentials; the residues automatically satisfy $v_c(0)=A+B$ and the correct initial slope.

Common pitfall. Getting the sign or placement of the initial-condition sources wrong (the inductor IC source is $i_L(0)/s$ in parallel; the capacitor IC source is $Cv_c(0)$), or forgetting that the final value is 0 here because the inductor shorts the node at dc.

Source. Sadiku, Fundamentals of Electric Circuits §16 (applications of the Laplace transform to circuit analysis).

Question 6 of 6: Laplace-domain RLC response

Question 6: Laplace-domain RLC response [10 + 10]