GATE 2025 IN Q.64: solution and answer

GATE 2025 · INQ.64 · 2 marks · NAT

A signal $V_{M} = 5 sin (π t /3)$ V is applied to the circuit consisting of a switch $S$ and capacitor $C = 0.1 μ F$ , as shown in the figure. The output $V_{x}$ of the circuit is fed to an ADC having an input impedance consisting of a $10 M Ω$ resistance in parallel with a $0.1 μ F$ capacitor. If $S$ is opened at $t = 0.5$ s, the value of $V_{x}$ at $t = 1.5$ s will be ________ V (rounded off to two decimal places).

Note: Assume all components are ideal.

GATE 2025 IN question 64: Left: a plot of V_M (in V) against t (in s). — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

ANSWER

$V_{x} (1.5 s) = 2.5 e^{- 0.5} \approx 1.52 V$ .

Official answer: 1.50 to 1.54Source: Official answer key, GATE 2025 IN (IIT Roorkee)

While $S$ is closed ( $t < 0.5$ s). The ideal source (zero internal resistance) is connected directly across $C$ and across the ADC input ( $10 M Ω ∥ 0.1 μ F$ ). Nothing sits between the source and the node, so the node voltage is forced to follow the source exactly: $V_{x} (t) = V_{M} (t)$ .
Voltage at the instant of switching. The argument of the sine is in radians: $V_{x} (0. 5^{-}) = 5 sin (\frac{π \times 0.5}{3}) = 5 sin \frac{π}{6} = 5 \times 0.5 = 2.5 V$ Both capacitors ( $C$ and the ADC's $0.1 μ F$ ) are charged to $2.5 V$ .
Continuity of capacitor voltage. A capacitor voltage cannot change instantly, so $V_{x} (0. 5^{+}) = V_{x} (0. 5^{-}) = 2.5 V$ . The two capacitors are at the same voltage when $S$ opens, so no charge moves between them.
Circuit for $t > 0.5$ s (source-free RC). With $S$ open the source is disconnected. What remains is $C = 0.1 μ F$ , the ADC's $0.1 μ F$ and the ADC's $10 M Ω$ , all in parallel between the node and GND. The two capacitors in parallel give $C_{e q} = 0.1 + 0.1 = 0.2 μ F$
Time constant. $τ = R C_{e q} = (10 \times 1 0^{6} Ω) (0.2 \times 1 0^{- 6} F) = 2 s$
Natural (exponential) decay, measured from the switching instant. $V_{x} (t) = 2.5 e^{- (t - 0.5) /2} V, t \geq 0.5 s$
At $t = 1.5$ s. $V_{x} (1.5) = 2.5 e^{- (1.5 - 0.5) /2} = 2.5 e^{- 0.5} = 2.5 \times 0.6065 = 1.516 V \approx 1.52 V$

KEY IDEA

While $S$ is closed, the ideal source holds the node at $V_{M} (t)$ . When $S$ opens at $t = 0.5$ s, the node starts at $5 sin (π /6) = 2.5 V$ and discharges through the ADC's $10 M Ω$ . The ADC's own $0.1 μ F$ is in parallel with $C$ , so $τ = R (C + C_{A D C}) = 10 M Ω \times 0.2 μ F = 2 s$ .

WATCH OUT FOR

The most common slip is leaving out the ADC's input capacitance. That gives $τ = 1 s$ and $V_{x} = 2.5 e^{- 1} \approx 0.92 V$ . Other slips:<br>(a) starting the decay from the peak value 5 V instead of the instantaneous value $5 sin (π /6) = 2.5 V$ ;<br>(b) evaluating $sin (0.5 π /3)$ as if the angle were in degrees;<br>(c) measuring the decay from $t = 0$ instead of from the switching instant, which gives $2.5 e^{- 1.5/2} \approx 1.18 V$ .

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 20 Transients questions, with notes

GATE 2025 IN Q.64: solution and answer

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