GATE 2025 EE Q.57: solution and answer

GATE 2025 · EEQ.57 · 2 marks · NAT

Q.57 The switch (S) closes at $t = 0$ sec. The time, in sec, the capacitor takes to charge to 50 V is _____ (round off to one decimal place).

GATE 2025 EE question 57: A single-loop circuit with a parallel group on its right. — Figure from the GATE 2025 EE paper. It is part of the question.

Answer and solution

ANSWER

The capacitor reaches 50 V at $t = 10 ln (3/2) \approx 4.1$ s.

Official answer: 4.0 to 4.2Source: Official answer key, GATE 2025 EE (IIT Roorkee)

Sign convention: take $v_{C}$ as the voltage of the capacitor plate on the source side minus the plate on node A's side.
Steady state before $t = 0$ (S open): at DC the capacitor is an open circuit, so no current flows in the series path through the 100 V source, the 5 F capacitor and the $2 Ω$ resistor. All of the 3 A therefore circulates through the current source and the $25 Ω$ resistor.
Voltage across the parallel group: $v_{A B} = 3 \times 25 = 75 V$ , with the top node A positive (the source pushes current up into A, and it returns down through the $25 Ω$ resistor). The $2 Ω$ resistor carries no current, so it drops no voltage.
KVL around the outer loop (source, capacitor, A to B, then the $2 Ω$ resistor): $100 = v_{C} (0^{-}) + 75 + 0$ , so $v_{C} (0^{-}) = 25 V$ . The capacitor is already partly charged.
Continuity of capacitor voltage: $v_{C} (0^{+}) = v_{C} (0^{-}) = 25 V$ .
After S closes: the switch shorts A to B, which removes the $25 Ω$ resistor and the 3 A source from the circuit. What remains is a series loop of the 100 V source, the 5 F capacitor and the $2 Ω$ resistor. So $v_{C} (\infty) = 100 V$ with the same polarity, and $τ = R C = 2 \times 5 = 10 s$ .
First-order response: $v_{C} (t) = v_{C} (\infty) + [v_{C} (0^{+}) - v_{C} (\infty)] e^{- t / τ} = 100 - 75 e^{- t /10} V$ .
Set $v_{C} = 50 V$ : $75 e^{- t /10} = 50 \Rightarrow e^{- t /10} = \frac{2}{3}$ .
$t = 10 ln \frac{3}{2} = 10 \times 0.4055 = 4.05 s \approx 4.1 s$

KEY IDEA

The capacitor does not start uncharged. Before the switch closes, the 3 A source drives 75 V across the 25 Ω resistor, and that voltage opposes the 100 V source, so $v_{C} (0) = 100 - 75 = 25$ V. Closing S shorts out the whole parallel group. That leaves a plain series RC loop (100 V, 5 F, 2 Ω) with $v_{C} (\infty) = 100$ V and $τ = 10$ s. Then use $v_{C} (t) = v_{C} (\infty) + [v_{C} (0^{+}) - v_{C} (\infty)] e^{- t / τ}$ .

WATCH OUT FOR

Assuming the capacitor starts at 0 V gives $v_{C} = 100 (1 - e^{- t /10})$ and $t = 10 ln 2 \approx 6.9$ s, which is wrong. Another trap is taking $v_{C} (0)$ as the 75 V across the parallel group rather than 100 − 75 = 25 V. A third is leaving the 25 Ω resistor in the time constant after S closes, when the switch has shorted it out.

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GATE 2025 EE Q.57: solution and answer

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