GATE 2024 EC Q.32: solution and answer

Official answer: 0.75Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. For $t>0$ the switch is closed. The circuit still has only one energy-storage element, the $3\text{H}$ inductor, so it is a first-order circuit with time constant $\tau =\frac{L}{R_{th}}$. Here $R_{th}$ is the Thevenin resistance seen from the inductor's terminals.

2. To find $R_{th}$, switch off the independent source: the $5\text{A}$ current source becomes an open circuit.

3. Left of the inductor: with the source open, the only path from the inductor's left terminal to ground is the $2\Omega$ resistor.

4. Right of the inductor: with S closed, the two $4\Omega$ resistors both run from the inductor's right terminal to ground, so they are in parallel: $4\parallel 4=\frac{4\times 4}{4+4}=2\Omega$.

5. Tracing a loop out of one inductor terminal and back into the other, the ground rail joins these two parts, so they are in series: $$R_{th}=2+2=4\Omega$$

6. Time constant: $$\tau =\frac{L}{R_{th}}=\frac{3}{4}=0.75\text{s}$$

7. The initial current, $i_L(0^{-})=5\times \frac{2}{2+4}=\frac{5}{3}\text{A}$ (the inductor acts as a short in the old steady state with S open), fixes where the transient starts. It does not affect $\tau$.