Question 6 of 6: Second-order RLC transient by Laplace
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, superposition, node analysis with dependent sources, first-order RC transients, AC phasor/node analysis, Thévenin equivalents and maximum-power transfer, and Laplace (s-domain) analysis of second-order circuits; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of 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; because $e$ and $i$ in Q5 are written as $40\sqrt2\cos$ and $5\sqrt2\cos$ (peak $=\sqrt2\times$rms), rms phasors $\mathbf E=40\angle30^\circ$ V and $\mathbf I=5\angle0^\circ$ A are used so the computed power is in watts directly.
Figure‑6 — series RLC. Switch on a (E1=20 V) for a long time, moved to b (E2=10 V) at t=0. i flows through R–L into C; Vc is measured across C.
Given. A series $R=5\,\Omega$, $L=2\text{ H}$, $C=\tfrac14\text{ F}$ loop is fed from $E_1=20\text{ V}$ (position-a) or $E_2=10\text{ V}$ (position-b). $V_c$ is measured across $C$ ($+$ on top) and $i$ is the loop current.
Element
Value
Source
Value
$R$
$5\,\Omega$
$E_1$ (pos. a)
$20\text{ V}$
$L$
$2\text{ H}$
$E_2$ (pos. b)
$10\text{ V}$
$C$
$\tfrac14\text{ F}$
Find. initial conditions $V_c(0^+),i(0^+)$; the s-domain circuit; and $V_c(t)$ for $t\ge0$.
Approach. Read the initial conditions from the position-a dc steady state (inductor short, capacitor open), transform the loop to the s-domain with initial-condition sources, solve for $V_c(s)$, and invert.
(a) Initial conditions. On position-a for a long time the capacitor is fully charged (open) and the inductor is a short with no loop current, so $$\boxed{i(0^+)=i_L(0^-)=0\text{ A},\qquad V_c(0^+)=V_c(0^-)=E_1=20\text{ V}.}$$ Both $i_L$ and $V_c$ are continuous across the switching.
(b) Laplace-transformed circuit. For $t\ge0$ the source is $E_2/s=10/s$. The elements become $R=5$, $sL=2s$ (series IC source $Li(0)=0$, hence none), and $1/(sC)=4/s$ in series with the initial-voltage source $V_c(0)/s=20/s$ (oriented to hold $+20\text{ V}$ across $C$ at $t=0$). The transformed loop is a single mesh: $10/s$ → $5$ → $2s$ → $\big[4/s$ with $20/s$ in series$\big]$ → back to the reference.
Mesh equation in $s$. KVL around the loop: $$\frac{10}{s}=\Big(2s+5+\frac4s\Big)\mathbf I(s)+\frac{20}{s}\;\Rightarrow\; \mathbf I(s)=\frac{-10}{2s^{2}+5s+4}.$$
Capacitor voltage in $s$. $V_c(s)=\dfrac{4}{s}\mathbf I(s)+\dfrac{20}{s}=\dfrac{20}{s}-\dfrac{40}{s(2s^{2}+5s+4)}=\dfrac{10}{s}+\dfrac{10s+25}{s^{2}+2.5s+2}.$
Complete the square. $s^{2}+2.5s+2=(s+1.25)^{2}+(0.6614)^{2}$, so $\alpha=1.25\text{ s}^{-1}$ and $\omega_d=0.6614\text{ rad/s}$ (under-damped). Writing $10s+25=10(s+1.25)+12.5$: $$V_c(s)=\frac{10}{s}+\frac{10(s+1.25)}{(s+1.25)^2+\omega_d^2}+\frac{12.5}{(s+1.25)^2+\omega_d^2}.$$
(c) Invert. Using the $e^{-at}\cos$ and $e^{-at}\sin$ pairs from the supplied table, $$\boxed{V_c(t)=10+e^{-1.25t}\big(10\cos0.6614t+18.90\sin0.6614t\big)\text{ V},\quad t\ge0.}$$ This starts at $V_c(0)=20\text{ V}$, ends at $V_c(\infty)=10\text{ V}=E_2$, and rings at $\omega_d=0.661\text{ rad/s}$ with envelope decay $\alpha=1.25\text{ s}^{-1}$.
