GATE 2018 EC Q.35: solution and answer

GATE 2018 · ECQ.35 · 2 marks · MCQ

For the circuit given in the figure, the voltage $V_{C}$ (in volts) across the capacitor is

GATE 2018 EC question 35: A single series loop. — Figure from the GATE 2018 EC paper. It is part of the question.

(A)
$1.25 2 sin (5 t - 0.25 π)$
(B)
$1.25 2 sin (5 t - 0.125 π)$
(C)
$2.5 2 sin (5 t - 0.25 π)$
(D)
$2.5 2 sin (5 t - 0.125 π)$

Answer and solution

ANSWER

$V_{C} (t) = 2.5 2 sin (5 t - 0.25 π) V$ , option (C).

Official answer: CSource: Official answer key, GATE 2018 EC (IIT Guwahati)

Combine the series resistors: there is only one loop, so both $100 k Ω$ resistors carry the same current as the capacitor. The one on the return wire counts just as much as the one on top. Together they act as $R = 200 k Ω$ in series with the capacitor.
Capacitor impedance at $ω = 5 rad/s$ : $Z_{C} = \frac{1}{j ω C} = \frac{1}{j ( 5 ) ( 1 \times 1 0 ^{- 6} )} = - j 200 k Ω$ .
Source phasor: take $sin (5 t)$ as the phase reference. The options are written as amplitudes, so we work with peak phasors: $V_{s} = 5∠ 0^{\circ} V$ (peak).
Voltage divider: the source's + terminal and the capacitor's + plate are both on the top rail, so the divider applies with no sign change: $V_{C} = V_{s} \frac{Z _{C}}{R + Z _{C}} = V_{s} \frac{1}{1 + j ω R C} .$
The key product: $ω R C = 5 \times (200 \times 1 0^{3}) \times (1 \times 1 0^{- 6}) = 1$ , so $V_{C} = \frac{5∠ 0 ^{\circ}}{1 + j} = \frac{5∠ 0 ^{\circ}}{2 ∠4 5 ^{\circ}} = \frac{5}{2} ∠ - 4 5^{\circ} V (peak) .$
Back to the time domain: $\frac{5}{2} = 2.5 2 \approx 3.54$ and $- 4 5^{\circ} = - 0.25 π$ . So $V_{C} (t) = 2.5 2 sin (5 t - 0.25 π) V$ : the capacitor voltage lags the source by $4 5^{\circ}$ . This is option (C).

KEY IDEA

This is a series RC voltage divider. Both $100 k Ω$ resistors sit in the one loop, so they add to $R = 200 k Ω$ no matter where they are drawn. With $ω R C = 1$ the divider ratio $\frac{1}{1 + j ω R C}$ becomes $\frac{1}{2} ∠ - 4 5^{\circ}$ : the amplitude drops by $2$ and the phase lags by $π /4$ .

WATCH OUT FOR

(1) Leaving out the resistor on the bottom return wire. That uses $R = 100 k Ω$ , so $ω R C = 0.5$ , giving about $4.47∠ - 26. 6^{\circ}$ , which is not among the options. (2) Converting to rms and then dividing by $2$ a second time. The rms value of $V_{C}$ is $2.5 V$ ; dividing again gives $1.25 2$ , which is the magnitude in options (A)/(B). (3) Halving the phase to $- 0.125 π$ , or using $ω = 2 π \times 5$ when the $5$ in $sin (5 t)$ is already $ω$ in rad/s.

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2018 EC Q.35: solution and answer

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