ISRO EC 2025 Q.37: solution and answer

Official answer: DSource: ISRO's final answer key of 15 May 2026

1. For $t<0$ the circuit is in steady state with the switch at 'a'. The capacitor carries no current, so the 15 kΩ and 10 kΩ resistors form a simple divider, and the capacitor sits across the lower 10 kΩ resistor. The magnitude of the voltage across the lower resistor is $75\cdot \frac{10}{15+10}=30\text{V}$.

2. Polarity: the + terminal of the 75 V source is on the bottom rail. The divider midpoint (the capacitor's marked + plate) is therefore 30 V below the bottom rail, so $V_c(0^{-})=-30\text{V}$. A capacitor voltage cannot change instantly, so $V_c(0^{+})=-30\text{V}$.

3. For $t>0$ the switch is at 'b': the source and divider are disconnected, and the capacitor is in series with the $20\text{k}\Omega$ resistor, across the parallel pair formed by the $5\text{mA}$ current source and the $5\text{k}\Omega$ resistor.

4. Final value: in steady state the capacitor carries no current, so no current flows in the $20\text{k}\Omega$ resistor and there is no drop across it. The whole $5\text{mA}$ flows (upward, into the top-right node) through the $5\text{k}\Omega$ resistor, so $V_c(\infty )=5\text{mA}\times 5\text{k}\Omega =+25\text{V}$.

5. Time constant: set the current source to zero (open circuit). The capacitor then sees $20\text{k}\Omega$ in series with $5\text{k}\Omega$, so $R_{th}=25\text{k}\Omega$.

6. $$\tau =R_{th}C=25\times 10^3\times 100\times 10^{-9}=2.5\times 10^{-3}\text{s}=2.5\text{ms}$$

7. The initial voltage is $-30\text{V}$, the final voltage is $25\text{V}$ and $\tau =2.5\text{ms}$, which is option (d). Options (a) and (b) need $\tau =25\text{ms}$, i.e. $250\text{k}\Omega$; options (a) and (c) need a final value of $125\text{V}$, i.e. $25\text{k}\Omega$ across the current source.