GATE 2026 EE Q.43: solution and answer

GATE 2026 · EEQ.43 · 2 marks · MCQ

The electrical network shown has an independent voltage source (10 V) and a current source (1 u(t) mA).

[Figure]

The voltage across the capacitor at time instants (in seconds) $t = 0^{+}$ , $t = 0.50$ , and $t = \infty$ , respectively, is:

GATE 2026 EE question 43: Single-node-pair circuit with a common bottom (reference) wire. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
8.00 V, 28.00 V, 26.36 V
(B)
8.00 V, 26.36 V, 28.00 V
(C)
10.00 V, 26.36 V, 28.00 V
(D)
10.00 V, 28.00 V, 26.36 V

Answer and solution

ANSWER

(B) 8.00 V, 26.36 V, 28.00 V: the capacitor starts at 8 V (the pre-step divider value) and rises towards 28 V with a time constant of 0.2 s, reaching 26.36 V at t = 0.5 s.

Official answer: BSource: Official answer key, GATE 2026 EE (IIT Guwahati)

Before the step ( $t < 0$ ), the capacitor is open at DC. The current source is zero, so the capacitor sits across the $100 k Ω$ arm of a voltage divider fed by the 10 V source through $25 k Ω$ : $v_{C} (0^{-}) = 10 \times \frac{100}{25 + 100} = 8 V$
Capacitor voltage cannot jump: $v_{C} (0^{+}) = v_{C} (0^{-}) = 8 V$ . This rules out options (C) and (D).
Norton equivalent seen by the capacitor for $t > 0$ . Source-transform the 10 V source and its series $25 k Ω$ into a current of $10/25 k Ω = 0.4 mA$ into the top node. It flows the same way as the 1 mA source (into the top node), so $I_{N} = 1.4 mA$ . The resistance is $R_{N} = 25 k Ω ∥ 100 k Ω = 20 k Ω$ .
Final value (the capacitor is open again at steady state): $v_{C} (\infty) = I_{N} R_{N} = 1.4 mA \times 20 k Ω = 28 V$
Time constant: $τ = R_{N} C = 20 k Ω \times 10 μ F = 0.2 s$ .
First-order response: $v_{C} (t) = v_{C} (\infty) + [v_{C} (0^{+}) - v_{C} (\infty)] e^{- t / τ} = 28 - 20 e^{- 5 t} V, t \geq 0$
At $t = 0.5$ s: $v_{C} (0.5) = 28 - 20 e^{- 2.5} = 28 - 20 (0.08208) = 28 - 1.642 = 26.36 V$
The three values are $8.00 V$ , $26.36 V$ and $28.00 V$ , which is option (B).

KEY IDEA

For a single capacitor, find three things: the initial value from the DC circuit before the switching event (capacitor open, and its voltage is continuous), the final value from the DC circuit after it, and $τ = R_{t h} C$ , where $R_{t h}$ is seen from the capacitor terminals with the independent sources killed. Then $v_{C} (t) = v_{C} (\infty) + [v_{C} (0^{+}) - v_{C} (\infty)] e^{- t / τ}$ . A source transformation makes both the final value and $R_{t h}$ one-line calculations here.

WATCH OUT FOR

Taking $v_{C} (0^{+}) = 10 V$ by assuming the capacitor charges to the full source voltage and forgetting the 100 kΩ divider arm. This leads to (C) or (D). The second trap is swapping the last two values (A/D). The response rises monotonically towards 28 V, so its value at 0.5 s (2.5 time constants) must be below 28 V. Some students also use only 100 kΩ or only 25 kΩ for $τ$ instead of their parallel combination, 20 kΩ.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 20 Transients questions, with notes

GATE 2026 EE Q.43: solution and answer

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