Second-order transient by the Laplace method: 22-Elec-A1 Circuits, Undated paper, Q6

Question 6: Second-order transient by the Laplace method [4 + 8 + 8]

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.

Figure 6 — position a connects the $1\text{ H}$–$0.1\text{ F}$ load to a $10\text{ A}\,\|\,5\,\Omega$ source; position b connects it instead to a $2\,\Omega$ to ground, giving a source-free series RLC ($2\,\Omega$, $1\text{ H}$, $0.1\text{ F}$) for $t\ge0$.

Given. $L=1\text{ H}$, $C=0.1\text{ F}$. For $t<0$ (pos a) the load hangs on the $10\text{ A}\,\|\,5\,\Omega$ source; for $t\ge0$ (pos b) it forms a series loop with a $2\,\Omega$ to ground and no source.

Find. $V_c(0^+),\,i_L(0^+),\,V_c(t)$.

Approach. Get the initial conditions from the pre-switch DC steady state, then write the natural (source-free) series-RLC response — conveniently via the Laplace transform with the initial-condition sources included.

Initial conditions (pos a, DC steady state). The inductor is a short and the capacitor an open. The open capacitor blocks any DC path through the $1\text{ H}$, so $\boxed{i_L(0^+)=0}$. With no inductor current, all $10\text{ A}$ flows through the $5\,\Omega$, and the inductor (a short) puts that node voltage on the capacitor: $\boxed{V_c(0^+)=10\times5=50\text{ V}}$.
Circuit for $t\ge0$. Position b isolates the source; the loop is a source-free series RLC: $R=2\,\Omega$, $L=1\text{ H}$, $C=0.1\text{ F}$, driven only by the stored energy $V_c(0^+)=50\text{ V}$, $i_L(0^+)=0$.
Laplace-transformed circuit (part b). Replace each element by its $s$-domain model with initial conditions: resistor $\to2$; inductor $\to sL=s$ in series with an IC source $L\,i_L(0^+)=0$; capacitor $\to \dfrac{1}{sC}=\dfrac{10}{s}$ in series with an IC voltage source $\dfrac{V_c(0^+)}{s}=\dfrac{50}{s}$. These form a single series loop.
Characteristic equation. The mesh gives $LC\,\ddot V_c+RC\,\dot V_c+V_c=0$, i.e. $$s^2+2s+10=0\ \Rightarrow\ s=-1\pm j3,$$ so the response is under-damped ($\alpha=1$, $\omega_d=3$, $\omega_0=\sqrt{10}$).
Solve $V_c(s)$. Transforming with $V_c(0)=50$, $\dot V_c(0)=i_L(0)/C=0$: $$V_c(s)=\frac{50s+100}{s^2+2s+10}=\frac{50(s+1)}{(s+1)^2+9}+\frac{50}{(s+1)^2+9}.$$
Invert. $$\boxed{V_c(t)=e^{-t}\big(50\cos3t+16.67\sin3t\big)\text{ V}=52.70\,e^{-t}\cos(3t-18.4^\circ)\text{ V},\ t\ge0.}$$

Quantity	Result
$V_c(0^+)$	$50\text{ V}$
$i_L(0^+)$	$0\text{ A}$
Natural frequencies	$s=-1\pm j3\ \text{(under-damped)}$
$V_c(t)$	$e^{-t}(50\cos3t+16.67\sin3t)\text{ V}$

Concept & Theory

Topic: the source-free series-RLC response via Laplace transforms.

Key relations: $s$-domain elements $R,\ sL,\ 1/sC$ with IC sources $L\,i_L(0)$ and $V_c(0)/s$; characteristic $s^2+\tfrac{R}{L}s+\tfrac{1}{LC}=0$; under-damped pair $e^{-\alpha t}(\cos\omega_d t,\ \sin\omega_d t)$ with $\alpha=R/2L$, $\omega_d=\sqrt{\omega_0^2-\alpha^2}$.

Why this works: the Laplace method turns the loop’s integro-differential KVL into an algebraic equation whose poles are the natural frequencies; the initial-condition sources inject the stored energy so the transform already contains the complete (natural) response.

Common pitfall: the final value — here the loop is source-free so $V_c(\infty)=0$ (no forced term); and $\dot V_c(0)=i_L(0)/C$, which is zero, not arbitrary. Source: Sadiku, 7th ed., §8.3 (source-free series RLC) and Ch. 16 (Laplace-transform circuit analysis).

Question 6 of 6: Second-order transient by the Laplace method

Question 6: Second-order transient by the Laplace method [4 + 8 + 8]