Question 3 of 6: First-order RC switching transient

Question 3: First-order RC switching transient [(4+4) + (6+6)]

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. Left branch: source $20u(t)\text{ V}$ in series with $5\,\Omega$; the current $i$ is that branch current. A $10\,\Omega$ and the $0.2\text{ F}$ capacitor (voltage $V_c$, $+$ up) both shunt the top node X to the bottom rail. The closed switch ties node X to a $15\text{ V}$ source ($+$ up) on the right.

Find. (a) $V_c(0^-),\,i(0^-)$; (b) closed-form $V_c(t),\,i(t)$ for $t>0$.

[Figure not reproduced: Figure 3 — redrawn. Node X is common to the $10\,\Omega$, the $0.2\text{ F}$ capacitor and (through the switch) the $15\text{ V}$ source; the $5\,\Omega$ ties X to the $20u(t)$ source. Opening the switch removes the $15\text{ V}$ branch. See the official exam paper.]

Approach. Use DC steady state on each side of the switching instant with the capacitor as an open circuit, enforce $V_c$ continuity at $t=0$, then write the single-time-constant response $V_c(t)=V_c(\infty)+[V_c(0^+)-V_c(\infty)]e^{-t/\tau}$.

1. (a) Initial state ($t<0$, switch closed). Here $20u(t)=0$, so the left source is a short and the $5\,\Omega$ merely ties X to ground through $0\text{ V}$. The ideal $15\text{ V}$ source, connected through the closed switch, clamps node X. At DC the capacitor is open, so $$V_c(0^-)=15\text{ V}=\boxed{15\text{ V}}.$$
2. Branch current $i(0^-)$. With X at $15\text{ V}$ and the left source at $0\text{ V}$, the $5\,\Omega$ carries $$i(0^-)=\frac{0-15}{5}=\boxed{-3\text{ A}},$$ i.e. $3\text{ A}$ flowing from node X back into the shorted source.
3. (b) Final state ($t\to\infty$, switch open). The $15\text{ V}$ branch is gone and $20u(t)=20\text{ V}$. With the capacitor open, X is the divider of $20\text{ V}$ across $5\,\Omega$ and $10\,\Omega$: $$V_c(\infty)=20\cdot\frac{10}{5+10}=13.33\text{ V}.$$
4. Time constant. Kill the source; the capacitor sees $5\,\Omega\|10\,\Omega$: $$R_{th}=\frac{5\cdot10}{15}=3.333\,\Omega,\qquad\tau=R_{th}C=3.333\times0.2=0.667\text{ s}.$$
5. Capacitor voltage for $t>0$. Continuity gives $V_c(0^+)=V_c(0^-)=15\text{ V}$, so with $1/\tau=1.5\ \text{s}^{-1}$ $$\boxed{V_c(t)=13.33+1.67\,e^{-1.5t}\ \text{V}},\quad t\ge0.$$
6. Branch current for $t>0$. $i=(20-V_c)/5$, hence $$i(t)=\frac{20-13.33-1.67e^{-1.5t}}{5}=\boxed{1.333-0.333\,e^{-1.5t}\ \text{A}}.$$ Check: $i(0^+)=1.0\text{ A}$ (jumps from $-3\text{ A}$ as the switch removes the $15\text{ V}$ clamp) and $i(\infty)=1.333\text{ A}$ ✓.