Second-order series RLC switching, Laplace-domain solution: 23-Mechatronics-A2 Circuits and Electronics, December 2019, Q6

Question 6: Second-order series RLC switching, Laplace-domain solution [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.

Quantity	Value
Source $E$ (switched in at $t=0$)	15 V dc
Series resistor $R$	4 Ω
Capacitor $C$, $v_c(0)=5$ V (+ at top)	0.2 F
Inductor $L$, $i_L(0)=1$ A (clockwise)	1 H

Find. The Laplace-domain circuit at $t\ge0$, and $V_c(t)$ for $t\ge0$.

Figure-6: 15 V dc source switched in at $t=0$, in series around a single loop with $R=4\,\Omega$, $C=0.2$ F ($v_c(0)=5$ V), and $L=1$ H ($i_L(0)=1$ A, same direction as the assumed loop current).

Approach. Replace $L$ and $C$ with their $s$-domain impedances plus series sources for the initial conditions, write one KVL loop equation for $I(s)$, then integrate to $V_c(s)$ and invert.

Laplace-domain circuit. $E/s$ in series with $R$, with $L$ replaced by $sL$ in series with a source $Li_L(0)=1\,\text{V}$ (opposing the assumed current direction), and $C$ replaced by $1/(sC)$ in series with a source $v_c(0)/s=5/s$ (aiding, matching its stated polarity): $$I(s)\Big[R+sL+\frac{1}{sC}\Big]=\frac{E}{s}-\frac{v_c(0)}{s}+Li_L(0).$$
Reduce to a single second-order ODE in $v_c(t)$. Using $i=C\,dv_c/dt$ eliminates $I(s)$ directly in the time domain: $$LC\,\ddot v_c+RC\,\dot v_c+v_c=E,\qquad v_c(0)=5\ \text{V},\quad \dot v_c(0)=\frac{i_L(0)}{C}=\frac{1}{0.2}=5\ \text{V/s}.$$ Substituting values: $$0.2\ddot v_c+0.8\dot v_c+v_c=15\ \Rightarrow\ \ddot v_c+4\dot v_c+5v_c=75.$$
Characteristic roots (Laplace denominator). $$s^2+4s+5=0\ \Rightarrow\ s=-2\pm j1\quad(\text{underdamped},\ \alpha=2,\ \omega_d=1\ \text{rad/s}).$$
Particular and homogeneous solution. Steady state $v_{c,p}=75/5=15$ V; homogeneous $v_{c,h}=e^{-2t}(A\cos t+B\sin t)$. Applying $v_c(0)=5\Rightarrow A=-10$ and $\dot v_c(0)=5\Rightarrow-2A+B=5\Rightarrow B=-15$.
Assemble $V_c(t)$. $$\boxed{V_c(t)=15-e^{-2t}\big(10\cos t+15\sin t\big)\ \text{V}},\quad t\ge0.$$ Check: $V_c(0)=15-10=5$ V and $\dot V_c(0)=-2(-10)+(-15)=5$ V/s, matching the given initial conditions.

Quantity	Result
Characteristic roots	$s=-2\pm j1$
$V_c(\infty)$	15 V
$V_c(t)$	$\boxed{15-e^{-2t}(10\cos t+15\sin t)\ \text{V}}$

Concept & Theory

Topic. Second-order series RLC step response via the Laplace transform.

Key relations. $\mathcal{L}\{L\,di/dt\}=sLI(s)-Li(0)$; $\mathcal{L}\{\frac{1}{C}\int i\,dt\}=\frac{I(s)}{sC}+\frac{v_c(0)}{s}$; characteristic equation $s^2+2\alpha s+\omega_0^2=0$ with $\alpha=R/2L$, $\omega_0^2=1/LC$.

Why this works. The Laplace transform converts each energy-storage element’s initial condition into an explicit series source in the $s$-domain, turning the integro-differential loop equation into ordinary algebra; the resulting characteristic polynomial roots (here complex, giving an underdamped sinusoid riding on the dc steady-state value) come directly from $R$, $L$, $C$ with no need to guess the form of the response in advance.

Common pitfall. Dropping the $v_c(0)$ series source or getting its polarity backwards relative to the assumed loop-current direction; forgetting that $\dot v_c(0)=i_L(0)/C$, not $i_L(0)$ itself.

Source. Hayt et al., Engineering Circuit Analysis §13.5–14.3 (Laplace circuit models, second-order responses).

Question 6 of 11: Second-order series RLC switching, Laplace-domain solution

Part A — Circuits

Question 6: Second-order series RLC switching, Laplace-domain solution [10, 10]