GATE 2023 EE Q.39: solution and answer

GATE 2023 · EEQ.39 · 2 marks · MCQ

A 3-phase, star-connected, balanced load is supplied from a 3-phase, 400 V (rms), balanced voltage source with phase sequence R-Y-B, as shown in the figure. If the wattmeter reading is $- 400 W$ and the line current is $I_{R} = 2 A$ (rms), then the power factor of the load per phase is

GATE 2023 EE question 39: Left: a box labelled '3 Phase 400 V Balanced Source' with three output lines labelled R (top), Y (middle) and B (bottom). — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
Unity
(B)
0.5 leading
(C)
0.866 leading
(D)
0.707 lagging

Answer and solution

ANSWER

The power factor per phase is 0.866 leading, option (C). The meter carries $I_{R}$ and sees $V_{Y B}$ , so it reads $V_{L} I_{L} sin ϕ = 800 sin ϕ = - 400 W$ . That gives $sin ϕ = - 0.5$ and $ϕ = - 3 0^{\circ}$ (leading).

Official answer: CSource: Official answer key, GATE 2023 EE (IIT Kanpur)

What the meter sees. The current coil carries $I_{R}$ , which enters at M. The pressure coil has its + end (C) on Y and its − end (V) on B, so it measures $V_{Y B}$ . The reading is $W = ∣ V_{Y B} ∣ ∣ I_{R} ∣ cos (∠ V_{Y B} - ∠ I_{R})$
Phasors for sequence R-Y-B. Take the star load's $V_{R N}$ as reference, with $V_{p} = 400/ 3 V$ (rms): $V_{R N} = V_{p} ∠ 0^{\circ}, V_{Y N} = V_{p} ∠ - 12 0^{\circ}, V_{B N} = V_{p} ∠12 0^{\circ}$
Line voltage across the pressure coil: $V_{Y B} = V_{Y N} - V_{B N} = V_{p} [(- 0.5 - j 0.866) - (- 0.5 + j 0.866)] = - j 3 V_{p} = 400∠ - 9 0^{\circ} V (rms)$
Line current. In a star load the line current equals the phase current, so $I_{R} = 2∠ - ϕ A (rms)$ . Here $ϕ$ is the angle by which the load current lags its own phase voltage $V_{R N}$ : $ϕ > 0$ means lagging and $ϕ < 0$ means leading.
Angle between the coil quantities: $∠ V_{Y B} - ∠ I_{R} = - 9 0^{\circ} + ϕ$ . Since $cos (ϕ - 9 0^{\circ}) = sin ϕ$ , the reading is $W = V_{L} I_{L} sin ϕ = 400 \times 2 \times sin ϕ = 800 sin ϕ$ So this connection measures reactive power, and $Q_{total} = 3 V_{p} I_{p} sin ϕ = 3 W$ .
Use the reading: $800 sin ϕ = - 400 \Rightarrow sin ϕ = - 0.5 \Rightarrow ϕ = - 3 0^{\circ}$ The current leads its phase voltage by $3 0^{\circ}$ , so the load is capacitive.
Power factor per phase: $cos 3 0^{\circ} = 0.866$ leading, which is option (C). Check: $Q_{total} = 3 \times (- 400) \approx - 693 var$ . The negative sign means the load supplies reactive power, which is what a leading (capacitive) load does.

KEY IDEA

With its current coil in one line (R) and its pressure coil across the other two lines (Y to B), one wattmeter reads $V_{L} I_{L} sin ϕ$ , not $V_{L} I_{L} cos ϕ$ . For sequence R-Y-B, $V_{Y B}$ lags $V_{R N}$ by exactly $9 0^{\circ}$ , so the reading measures reactive power: $Q_{total} = 3 W$ . The sign of the reading tells you whether the load lags or leads.

WATCH OUT FOR

There are three common mistakes. (1) Using the phase voltage $400/ 3 \approx 231 V$ instead of the line voltage across the pressure coil. That gives $sin ϕ = - 0.866$ and the trap option (B), 0.5 leading. (2) Reading the pressure coil as $V_{B Y}$ instead of $V_{Y B}$ (ignoring the + mark on C), or taking the sequence as R-B-Y. Either one flips the sign and gives 0.866 lagging. (3) Treating the reading as $V_{L} I_{L} cos ϕ$ as if it were an active-power meter.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 11 Three-phase (EE) questions, with notes

GATE 2023 EE Q.39: solution and answer

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