GATE 2018 IN Q.31: solution and answer

GATE 2018 · INQ.31 · 2 marks · NAT

In the figure, an RLC load is supplied by a 230 V, 50 Hz single phase source. The magnitude of the reactive power (in VAr) supplied by the source is ________.

GATE 2018 IN question 31: A 230 V, 50 Hz sinusoidal AC source sits on the left. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

ANSWER

The source supplies reactive power of magnitude about 67.33 VAr. The net Q is −67.33 VAr because the load is net capacitive.

Official answer: 66 to 68Source: Official answer key, GATE 2018 IN (IIT Guwahati)

Both branches are connected directly across the source, so each one sees the full $V = 230 V$ (rms). Complex power is conserved, so the reactive power the source supplies equals the sum of the reactive powers taken by the two branches. Work out each branch separately and add.
R-L branch impedance: $Z_{1} = 162.6 + j 162.6 = 162.6 2 ∠4 5^{\circ} \approx 229.95∠4 5^{\circ} Ω$ .
Branch current: $∣ I_{1} ∣ = \frac{230}{162.6 2} \approx 1.0002 A$ (rms).
Reactive power absorbed by the inductor: $Q_{L} = ∣ I_{1} ∣^{2} X_{L} = \frac{23 0 ^{2} \times 162.6}{2 \times 162. 6 ^{2}} = \frac{52900}{325.2} \approx 162.67 VAr (absorbed, lagging)$
Capacitor branch: $Q_{C} = \frac{V ^{2}}{X _{C}} = \frac{23 0 ^{2}}{230} = 230 VAr$ . The capacitor supplies this reactive power (leading).
Net reactive power supplied by the source: $Q = Q_{L} - Q_{C} = 162.67 - 230 = - 67.33 VAr$ . The negative sign means the load as a whole is capacitive, so the source actually absorbs this reactive power instead of supplying it.
Admittance cross-check: $Y = \frac{1}{162.6 ( 1 + j )} + \frac{1}{- j 230} = \frac{1}{325.2} + j (\frac{1}{230} - \frac{1}{325.2}) S$ . Then $S = V^{2} Y^{*} = 162.67 - j 67.33 VA$ , which gives the same $Q = - 67.33 VAr$ .
The question asks for the magnitude: $∣ Q ∣ \approx 67.33 VAr$ .

KEY IDEA

In parallel branches across one voltage source, reactive power adds branch by branch (complex power is conserved). The inductor absorbs $∣ I_{1} ∣^{2} X_{L} = V^{2} X_{L} /∣ Z_{1} ∣^{2}$ and the capacitor supplies $V^{2} / X_{C}$ . Their difference is what the source must provide or absorb.

WATCH OUT FOR

Writing $Q_{L} = V^{2} / X_{L} = 23 0^{2} /162.6 \approx 325.4 VAr$ , as if the inductor had the full 230 V across it. It is in series with the resistor, so only part of the voltage appears across it, and you must use $∣ I_{1} ∣^{2} X_{L}$ with $∣ I_{1} ∣ = V /∣ Z_{1} ∣$ . That shortcut gives $∣ Q ∣ \approx 95.4 VAr$ , which is wrong. A second slip is to report $- 67.33$ when the question asks for a magnitude.

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.31: solution and answer

Answer and solution

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