Laplace-domain transient analysis: 22-Elec-A1 Circuits, December 2013, Q6

Question 6: Laplace-domain transient analysis [10 + 5 + 5]

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.

Element	Value
DC source	10 V
Series resistor (source side)	1 Ω
Capacitor	1 F, $V_c(0)=8$ V
Inductor branch resistor	1 Ω
Inductor	1 H, $i_L(0)=4$ A

Find. The $s$-domain model, $I_L(s)$, and $i_L(t)$.

Figure 6 — Time-domain network. Node A carries the capacitor to ground and the $(1\,\Omega+1\,\text{H})$ branch to ground; $i_L(t)$ is the downward inductor current.

Approach. Replace each element by its $s$-domain model carrying its initial condition (source $10/s$; capacitor $1/s$ in series with $8/s$; inductor $s$ in series with $Li_L(0)=4$ V), then take one node equation and invert.

Figure 6(a) — Laplace-transformed circuit. The capacitor becomes $\tfrac1s$ with a series $\tfrac8s$ source; the inductor becomes $s$ with a series 4 V (=$Li_L(0)$) source; the DC source becomes $10/s$.

Node equation at A. With $V(s)$ the node voltage, the cap current leaving is $sV-8$ and the inductor-branch current is $(V+4)/(1+s)$: $$\frac{10/s-V}{1}=(sV-8)+\frac{V+4}{1+s}.$$
Solve for $V(s)$. Multiplying by $(1+s)$ and collecting terms gives $$V(s)=\frac{8s^{2}+14s+10}{s\,(s^{2}+2s+2)}.$$
Inductor current. $I_L(s)=\dfrac{V+4}{1+s}$; the numerator carries a factor $(s+1)$ that cancels the branch pole: $$I_L(s)=\boxed{\dfrac{4s^{2}+12s+10}{s\,(s^{2}+2s+2)}}.$$
Check the extremes. Initial/final-value theorems give $i_L(0^{+})=\lim_{s\to\infty}sI_L=4$ A and $i_L(\infty)=\lim_{s\to0}sI_L=\tfrac{10}{2}=5$ A — matching the 4 A initial current and the 10 V/(1+1) Ω DC steady state.
Partial fractions. $$I_L(s)=\frac{5}{s}+\frac{-s+2}{s^{2}+2s+2}=\frac{5}{s}+\frac{-(s+1)+3}{(s+1)^{2}+1}.$$
Invert. Using $\dfrac{s+a}{(s+a)^2+\omega^2}\!\to e^{-at}\cos\omega t$ and $\dfrac{\omega}{(s+a)^2+\omega^2}\!\to e^{-at}\sin\omega t$ (here $a=\omega=1$): $$\boxed{i_L(t)=5-e^{-t}\cos t+3e^{-t}\sin t\ \text{A},\quad t\ge0.}$$

Quantity	Result
$I_L(s)$	$\dfrac{4s^{2}+12s+10}{s(s^{2}+2s+2)}$
$i_L(0^{+})$ / $i_L(\infty)$	4 A / 5 A
$i_L(t)$	$\boxed{5-e^{-t}\cos t+3e^{-t}\sin t\ \text{A}}$

Concept & Theory

Topic. Laplace-domain (operational) analysis of a second-order circuit with stored initial energy.

Key relations. Capacitor: $I=sC\,V-C\,V_c(0)$ (equivalently $1/sC$ in series with $V_c(0)/s$). Inductor: $V=sL\,I-L\,i_L(0)$ (i.e. $sL$ in series with $Li_L(0)$). Initial-value $\lim_{s\to\infty}sF(s)$; final-value $\lim_{s\to0}sF(s)$.

Why this works. The $s$-domain models fold the initial conditions into equivalent sources, turning integro-differential KCL/KVL into algebra. The complex-conjugate poles at $s=-1\pm j1$ produce a damped oscillation $e^{-t}(\ldots)$ riding on the 5 A DC forced response set by the pole at the origin.

Common pitfall. Wrong sign or magnitude on an initial-condition source ($Li_L(0)$, not $i_L(0)/s$, for the series model), or forgetting the pole-at-origin term that carries the DC steady state.

Source. Sadiku §16.3–16.5 (circuit analysis in the $s$-domain with initial conditions); Hayt §14 (Laplace circuit models).

Question 6 of 6: Laplace-domain transient analysis

Question 6: Laplace-domain transient analysis [10 + 5 + 5]