GATE EC 2021 Set 1 — Question 38

The solution involves a two-step process: finding the initial conditions at $t=0^-$ and then analyzing the circuit at $t=0^+$.Step 1: Analyze the circuit for $t < 0$ (finding initial conditions)
For $t < 0$, the switch has been in position P for a long time. The circuit is in DC steady state. In DC steady state, a capacitor acts as an open circuit and an inductor acts as a short circuit.The capacitor is in parallel with the 20 kΩ resistor. The voltage across the capacitor, $v(t)$, will be the same as the voltage across the 20 kΩ resistor. The 5 kΩ and 20 kΩ resistors form a voltage divider with the 20 V source.$$v(0^-) = 20 \text{ V} \times \frac{20 \text{ k}\Omega}{5 \text{ k}\Omega + 20 \text{ k}\Omega} = 20 \times \frac{20}{25} = 16 \text{ V}$$Since the voltage across a capacitor cannot change instantaneously, $v(0^+) = v(0^-) = 16$ V.The inductor is in the part of the circuit that is disconnected for $t<0$. Therefore, there is no current flowing through it.$$i_L(0^-) = 0 \text{ A}$$Since the current through an inductor cannot change instantaneously, $i_L(0^+) = i_L(0^-) = 0$ A.Step 2: Analyze the circuit for $t > 0$
At $t=0$, the switch moves to position Q. The 20 V source and the 20 kΩ resistor are disconnected. The capacitor starts to discharge through a circuit consisting of a 5 kΩ resistor in parallel with a series combination of a 10 kΩ resistor and a 0.1 mH inductor.We need to find $\frac{dv(t)}{dt}$ at $t=0^+$. The current through the capacitor is given by $i_C(t) = C \frac{dv(t)}{dt}$. Therefore, $\frac{dv(0^+)}{dt} = \frac{i_C(0^+)}{C}$.Let's apply Kirchhoff's Current Law (KCL) at the top node (the positive terminal of the capacitor) at $t=0^+$. The current leaving the capacitor, $i_C(0^+)$, splits into two paths:KCL equation: $i_C(0^+) + i_{5k}(0^+) + i_L(0^+) = 0$.We can find the currents in the resistive and inductive branches at $t=0^+$:$$i_{5k}(0^+) = \frac{v(0^+)}{5 \text{ k}\Omega} = \frac{16 \text{ V}}{5 \times 10^3 \Omega} = 3.2 \text{ mA}$$$$i_L(0^+) = 0 \text{ A} \text{ (from initial conditions)}$$Now, we can find the capacitor current at $t=0^+$:$$i_C(0^+) = - (i_{5k}(0^+) + i_L(0^+)) = - (3.2 \text{ mA} + 0) = -3.2 \text{ mA}$$Finally, we can calculate $\frac{dv(0^+)}{dt}$:$$\frac{dv(0^+)}{dt} = \frac{i_C(0^+)}{C} = \frac{-3.2 \times 10^{-3} \text{ A}}{1 \text{ mF}} = \frac{-3.2 \times 10^{-3} \text{ A}}{1 \times 10^{-3} \text{ F}} = -3.2 \text{ V/s}$$The calculated value is -3.2 V/s. The closest option is (C) -3 V/s. This suggests a possible typo in the problem's component values or options, but -3 V/s is the intended answer.