GATE 2022 IN Q.60: solution and answer

GATE 2022 · INQ.60 · 2 marks · NAT

A 1 kHz sine-wave generator having an internal resistance of 50 Ω generates an open-circuit voltage of 10 V $_{p-p}$ . When a capacitor is connected across the output terminals, the voltage drops to 8 V $_{p-p}$ . The capacitance of the capacitor (in microfarads) is _______ (round off to two decimal places)

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Answer and solution

ANSWER

The capacitance is about 2.39 µF (X_C = 66.67 Ω at 1 kHz).

Official answer: 2.35 to 2.43Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Model the loaded generator. Treat the generator as an ideal sine source of $10 V_{p-p}$ at $f = 1 kHz$ in series with its internal resistance $R = 50 Ω$ . Connecting $C$ across the terminals makes a series $R$ – $C$ divider, and the terminal voltage is the voltage across $C$ .
Voltage divider in phasor form. $\frac{V _{C}}{V _{s}} = \frac{- j X _{C}}{R - j X _{C}} \Rightarrow \frac{V _{C}}{V _{s}} = \frac{X _{C}}{R ^{2} + X _{C}^{2}}$ Both voltages are sinusoids of the same frequency, so their peak-to-peak values are in the same ratio as their magnitudes: the ratio is $\frac{8}{10} = 0.8$ .
Solve for $X_{C}$ . $\frac{X _{C}^{2}}{R ^{2} + X _{C}^{2}} = 0.64 \Rightarrow 0.36 X_{C}^{2} = 0.64 R^{2} \Rightarrow X_{C} = \frac{0.8}{0.6} R = \frac{4}{3} (50) = 66.67 Ω$
Check with the phasor triangle. $V_{R}$ and $V_{C}$ are $9 0^{\circ}$ apart and add to the source voltage, so $V_{R} = 1 0^{2} - 8^{2} = 6 V_{p-p}$ . The same current flows through both, so $X_{C} / R = V_{C} / V_{R} = 8/6$ , which again gives $X_{C} = 66.67 Ω$ .
Capacitance from reactance. $C = \frac{1}{2 π f X _{C}} = \frac{1}{2 π ( 1000 ) ( 66.67 )} = \frac{1}{4.1888 \times 1 0 ^{5}} = 2.387 \times 1 0^{- 6} F$
Answer. $C \approx 2.39 μ F$ .

KEY IDEA

The internal resistance and the capacitor form a series R–C divider. V_R and V_C are in quadrature, so the voltages obey a right-angle (Pythagoras) relation: V_R = sqrt(10² − 8²) = 6 V p-p. Since the same current flows through both, X_C/R = 8/6. That gives X_C = 66.67 Ω, and C = 1/(2πf X_C).

WATCH OUT FOR

Treating the divider as purely resistive, i.e. 8/10 = X_C/(R + X_C). That gives X_C = 200 Ω and C ≈ 0.80 µF, because it ignores the 90° phase difference between V_R and V_C. A second mistake is taking the "missing" 2 V p-p as the resistor drop. The drops add as phasors, not as numbers, so the resistor drop is 6 V p-p.

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GATE 2022 IN Q.60: solution and answer

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