GATE EE 2023 Set 1 — Question 54

Reading the circuit. The top rail joins the $10\ \Omega$ resistor, the current source $i(t)=\sin 1000t$, switch $K$ and the top plate of the $100\ \mu$ F capacitor. The bottom of the resistor, the current source and $K$ share one node $N$; the lower plate of the capacitor is node $M$. Switch $\bar{K}$ joins $N$ to $M$, and the branch ($10\ \Omega$ in series with the 5 V battery, $+$ towards $N$) is connected between $N$ and $M$, in parallel with $\bar{K}$.1. Before switching ($t < t_1$): $K$ open, $\bar{K}$ closed.
$\bar{K}$ shorts the battery branch, so the capacitor sits directly across the $10\ \Omega$ resistor and the current source. In sinusoidal steady state$$Z = R \parallel \frac{1}{j\omega C} = \frac{R}{1 + j\omega RC} = \frac{10}{1 + j(1000)(10)(100\times10^{-6})} = \frac{10}{1+j} = 5\sqrt{2}\angle -45^\circ\ \Omega$$$$v_C(t) = 1\times 5\sqrt{2}\,\sin(1000t - 45^\circ)\ \text{V} \qquad (t < t_1)$$2. After switching ($t \ge t_1$): $K$ closed, $\bar{K}$ open.
$K$ shorts the resistor and the current source (top rail tied to $N$), and with $\bar{K}$ open the capacitor is now in series with the $10\ \Omega$ resistor and the 5 V battery. This is a first-order RC circuit with $\tau = RC = 1$ ms whose steady-state (final) capacitor voltage is$$v_C(\infty) = 5\ \text{V}$$and its response is $v_C(t) = 5 + [v_C(t_1) - 5]\,e^{-(t-t_1)/\tau}$.3. Condition for no transient.
The exponential term vanishes only if the capacitor voltage at the switching instant already equals the final value:$$5\sqrt{2}\,\sin(1000t_1 - 45^\circ) = 5 \implies \sin(1000t_1 - 45^\circ) = \frac{1}{\sqrt{2}}$$4. Smallest $t_1 > 0$.
The first solution is $1000t_1 - 45^\circ = 45^\circ$, i.e. $1000t_1 = 90^\circ = \pi/2$ rad:$$t_1 = \frac{\pi}{2000}\ \text{s} = 1.5708\times10^{-3}\ \text{s}$$Answer: $t_1 \approx 1.57$ ms