GATE 2022 IN Q.62: solution and answer

GATE 2022 · INQ.62 · 2 marks · NAT

In the circuit shown, the capacitance $C_{0} = 10 μ F$ and inductance $L_{0} = 1 mH$ and the diode is ideal. The capacitor is initially charged to 10 V and the current in the inductor is initially zero. If the switch is closed at $t = 0$ s, the voltage $V_{c} (t)$ (in volts) across the capacitor at $t = 0.5$ s is ______ (round off to one decimal place)

GATE 2022 IN question 62: A single series loop. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

ANSWER

$V_{c} (0.5 s) = - 10.0 V$ . The diode lets through exactly one half-cycle of the LC oscillation, which ends at $t = π / ω_{0} \approx 0.314 ms$ . That leaves the capacitor charged to the reversed polarity, and it stays there.

Official answer: -10.1 to -9.9Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Which way the first current flows. The capacitor's + plate is at the top. When the switch closes, current leaves the + plate, passes through the switch and then through the diode from anode to cathode, and returns down through $L_{0}$ . That is the diode's forward direction, so the diode conducts and the circuit is a series LC loop.
Natural frequency of the loop. $ω_{0} = \frac{1}{L _{0} C _{0}} = \frac{1}{( 1 0 ^{- 3} ) ( 10 \times 1 0 ^{- 6} )} = \frac{1}{1 0 ^{- 8}} = 1 0^{4} rad/s$
Solution while the diode conducts. The initial conditions are $v_{C} (0^{+}) = 10 V$ and $i_{L} (0^{+}) = i_{L} (0^{-}) = 0$ . For a lossless LC loop this gives $v_{C} (t) = 10 cos ω_{0} t V, i (t) = - C_{0} \frac{d v _{C}}{d t} = C_{0} V_{0} ω_{0} sin ω_{0} t = (1 0^{- 5}) (10) (1 0^{4}) sin (1 0^{4} t) = sin (1 0^{4} t) A$ Here $i$ is positive in the diode's forward direction, i.e. leaving the + plate.
The diode turns off. The current $i (t) = sin (1 0^{4} t)$ is positive for the first half-cycle and falls back to zero at $t_{1} = \frac{π}{ω _{0}} = \frac{π}{1 0 ^{4}} \approx 0.314 ms$ After $t_{1}$ the current would have to reverse, and an ideal diode cannot carry reverse current. So the diode turns off at $t_{1}$ and the loop stays open.
Capacitor voltage at turn-off. $v_{C} (t_{1}) = 10 cos π = - 10 V$ The charge has swung all the way across and the polarity is reversed. The inductor current is zero at this moment, so $L_{0}$ stores no energy. As a check, all of the initial energy $\frac{1}{2} C_{0} (10)^{2}$ is back in the capacitor, which is what a lossless swing should give.
The voltage is held from then on. With the loop open, the ideal (lossless) capacitor keeps its charge. The top of $C_{0}$ (the anode) is at $- 10 V$ relative to the bottom wire. The cathode connects to the bottom wire through $L_{0}$ , which carries no current and so has no voltage across it. The diode is therefore reverse biased by 10 V and stays off. Since $t = 0.5 s ≫ 0.314 ms$ , $V_{c} (0.5 s) = - 10.0 V$

KEY IDEA

A diode in series with a lossless LC loop lets current flow for only one half-cycle. The current $i = C_{0} V_{0} ω_{0} sin ω_{0} t$ returns to zero at $t = π / ω_{0}$ . At that instant the capacitor voltage has swung from $+ V_{0}$ to $- V_{0}$ . The diode then blocks the reverse current and the loop is open, so the capacitor holds $- V_{0}$ for ever after. Any time later than $π / ω_{0}$ (here 0.314 ms, far shorter than 0.5 s) gives $V_{c} = - V_{0}$ .

WATCH OUT FOR

One common mistake is to ignore the diode and evaluate the free LC oscillation at 0.5 s: $10 cos (1 0^{4} \times 0.5) = 10 cos (5000) \approx 1.5 V$ . Another is to assume the capacitor fully discharges and answer 0 V. A third is to forget that the charge reverses and answer $+ 10$ V. The diode stops the current at its first zero crossing, which comes when the capacitor is at $- 10$ V, so the sign of the answer matters.

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 20 Transients questions, with notes

GATE 2022 IN Q.62: solution and answer

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