GATE 2018 IN Q.8: solution and answer

GATE 2018 · INQ.8 · 1 mark · NAT

A coil having an impedance of $(10 + j 100) Ω$ is connected in parallel to a variable capacitor as shown in figure. Keeping the excitation frequency unchanged, the value of the capacitor is changed so that parallel resonance occurs. The impedance across terminals p-q at resonance (in $Ω$ ) is ______.

GATE 2018 IN question 8: Two open terminals, p at the top left and q at the bottom left. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

ANSWER

At parallel resonance the impedance across p-q is purely resistive: $Z_{pq} = \frac{R ^{2} + X _{L}^{2}}{R} = \frac{1 0 ^{2} + 10 0 ^{2}}{10} = 1010 Ω$ .

Official answer: 1009 to 1011Source: Official answer key, GATE 2018 IN (IIT Guwahati)

Frequency is fixed, so the coil stays at $Z_{L} = 10 + j 100 Ω$ . Only the capacitor's susceptance $B_{C} = ω C$ changes.
Admittance of the coil: $Y_{L} = \frac{1}{10 + j 100} = \frac{10 - j 100}{1 0 ^{2} + 10 0 ^{2}} = \frac{10}{10100} - j \frac{100}{10100} S$
Parallel resonance means the total admittance $Y = Y_{L} + j ω C$ is purely real, so the capacitor has to cancel the coil's susceptance: $ω C = \frac{100}{10100} S$ .
At resonance only the coil's conductance is left: $Y = \frac{10}{10100} S$ .
Impedance across p-q: $Z_{pq} = \frac{1}{Y} = \frac{10100}{10} = 1010 Ω$
Check with the dynamic-resistance form: $Z = \frac{R ^{2} + X _{L}^{2}}{R} = R (1 + Q^{2})$ with $Q = X_{L} / R = 10$ , which gives $10 \times 101 = 1010 Ω$ , purely resistive.

KEY IDEA

A lossy coil (R in series with L) in parallel with a capacitor resonates when the capacitor's susceptance cancels the coil's susceptance. This is the unity-power-factor condition: total admittance purely real. Only the coil's conductance $R / (R^{2} + X_{L}^{2})$ is left, so the circuit looks like the dynamic resistance $Z_{D} = (R^{2} + X_{L}^{2}) / R = R (1 + Q^{2})$ . For a high-Q coil this is roughly $L / (C R)$ .

WATCH OUT FOR

Treating resonance as $X_{C} = X_{L} = 100 Ω$ and then combining $(10 + j 100)$ with $- j 100$ in parallel gives $1000 - j 100 Ω$ . That is not purely resistive, so it is not resonance. With the coil resistance present, the capacitor has to cancel the coil's susceptance ( $ω C = 100/10100$ S), not its reactance. A second slip is answering $10 Ω$ , as if the circuit were series resonant: at parallel resonance the impedance is at its largest, not its smallest.

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

GATE 2018 IN Q.8: solution and answer

Answer and solution

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