GATE 2023 EE Q.54: solution and answer

GATE 2023 · EEQ.54 · 2 marks · NAT

The circuit shown in the figure is initially in the steady state with the switch $K$ in open condition and $\overline{K}$ in closed condition. The switch $K$ is closed and $\overline{K}$ is opened simultaneously at the instant $t = t_{1}$ , where $t_{1} > 0$ . The minimum value of $t_{1}$ in milliseconds, such that there is no transient in the voltage across the $100 μ F$ capacitor, is ________ (Round off to 2 decimal places).

GATE 2023 EE question 54: Three nodes: a top rail T, a bottom-left node M (drawn as a dot below switch K), and a bottom-right node R (a dot below… — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

The minimum switching instant is $t_{1} = \frac{π}{2} ms \approx 1.57 ms$ .

Official answer: 1.56 to 1.58Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Circuit before $t_{1}$ . With $\overline{K}$ closed, M and R are the same node, so the capacitor is in parallel with the left $10 Ω$ resistor and the current source. The battery branch is shorted by $\overline{K}$ : a current of $\frac{5}{10} = 0.5 A$ circulates through the battery, the lower resistor and $\overline{K}$ , and it has no effect on the capacitor. Take $v_{C} = v_{T} - v_{R}$ .
Sinusoidal steady state for $t < t_{1}$ . With $ω = 1000 rad/s$ : $ω C = 1000 \times 100 \times 1 0^{- 6} = 0.1 S$ and $G = \frac{1}{10} = 0.1 S$ , so $Y = 0.1 + j 0.1 S$ and $Z = \frac{1}{Y} = 5 - j 5 = 52 ∠ - 4 5^{\circ} Ω$
The source pushes current up into T, and it returns to M through the RC pair. With the peak phasor $I = 1∠ 0^{\circ} A$ , $V_{C} = I Z = 52 ∠ - 4 5^{\circ} V$ (peak). So for $t < t_{1}$ : $v_{C} (t) = 52 sin (1000 t - \frac{π}{4}) V$
Circuit after $t_{1}$ . Closing $K$ shorts T to M, so the current source and the left $10 Ω$ resistor no longer affect the capacitor (the source current just flows through $K$ ). With $\overline{K}$ open, the loop is: capacitor (T to R), then the battery and the lower $10 Ω$ resistor (R back to M, which is now T). This is a first-order RC circuit with $τ = R C = 10 \times 100 μ F = 1 ms$ .
Final value. In DC steady state no current flows, so there is no drop across the lower resistor, and $v_{C} (\infty) = v_{M} - v_{R} = + 5 V$ , because the battery's + terminal faces M and its $-$ terminal faces R.
Complete response. For $t \geq t_{1}$ : $v_{C} (t) = 5 + [v_{C} (t_{1}) - 5] e^{- (t - t_{1}) / τ}$ The capacitor voltage cannot jump, so $v_{C} (t_{1}^{+}) = v_{C} (t_{1}^{-})$ . The transient term disappears only if $v_{C} (t_{1}^{-}) = 5 V$ .
Set $52 sin (1000 t_{1} - \frac{π}{4}) = 5$ , which gives $sin (1000 t_{1} - \frac{π}{4}) = \frac{1}{2}$ . The solutions are $1000 t_{1} - \frac{π}{4} = \frac{π}{4}$ or $\frac{3 π}{4}$ , plus any multiple of $2 π$ . In other words, $1000 t_{1} = \frac{π}{2}$ or $π$ , plus any multiple of $2 π$ .
Since $t_{1} > 0$ , the smallest root is $1000 t_{1} = \frac{π}{2}$ . The other roots are either negative or larger than this. So $t_{1} = \frac{π}{2} ms = 1.5708 ms \approx 1.57 ms$

KEY IDEA

If the post-switching forced response is DC, the capacitor shows no transient only when you switch at the moment its pre-switching voltage already equals the new final value. Here that means the AC steady-state voltage $52 sin (1000 t - π /4)$ has to equal $+ 5 V$ . The time constant plays no part.

WATCH OUT FOR

One slip is to include the battery in the pre-switching circuit, but $\overline{K}$ shorts that branch. Another is to forget the $- 4 5^{\circ}$ phase of the parallel RC impedance and solve $52 sin 1000 t_{1} = 5$ , which gives $0.785 ms$ . A third is to misread the battery's polarity. Taking $v_{C} (\infty) = - 5 V$ leads to $4.71 ms$ .

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GATE 2023 EE Q.54: solution and answer

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