GATE 2020 IN Q.42: solution and answer

GATE 2020 · INQ.42 · 2 marks · NAT

The real power drawn by a balanced load connected to a 400 V, 50 Hz, balanced, symmetrical 3-phase, 3-wire, RYB sequence mains is measured using the two-wattmeter method. Wattmeter $W_{1}$ is connected in the R line and wattmeter $W_{2}$ is connected in the B line. The line current is measured as $\frac{1}{3}$ A. If the wattmeter $W_{1}$ reads zero, the reading on $W_{2}$ (in watts) is ____

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Answer and solution

ANSWER

The reading on $W_{2}$ is $200 W$ .

Official answer: 199 to 201Source: Official answer key, GATE 2020 IN (IIT Delhi)

Set up the connection. In the two-wattmeter method, the pressure coils of both meters return to the third line, Y. The current coil of $W_{1}$ is in R and the current coil of $W_{2}$ is in B. So $W_{1}$ measures $V_{R Y}$ against $I_{R}$ , and $W_{2}$ measures $V_{B Y}$ against $I_{B}$ .
Write the phasors for RYB sequence, taking $V_{R N} = V_{p} ∠ 0^{\circ}$ as reference, with $V_{Y N} = V_{p} ∠ - 12 0^{\circ}$ and $V_{B N} = V_{p} ∠12 0^{\circ}$ . Let the load power-factor angle be $ϕ$ , with lagging taken as positive: $V_{R Y} V_{B Y} = V_{R N} - V_{Y N} = 3 V_{p} ∠3 0^{\circ}, = V_{B N} - V_{Y N} = 3 V_{p} ∠9 0^{\circ}, I_{R} I_{B} = I_{L} ∠ - ϕ = I_{L} ∠ (12 0^{\circ} - ϕ)$
Wattmeter readings, using $W = V I cos (angle between V and I)$ with $V_{L} = 3 V_{p}$ : $W_{1} W_{2} = V_{L} I_{L} cos (3 0^{\circ} - (- ϕ)) = V_{L} I_{L} cos (3 0^{\circ} + ϕ) = V_{L} I_{L} cos (9 0^{\circ} - (12 0^{\circ} - ϕ)) = V_{L} I_{L} cos (3 0^{\circ} - ϕ)$
Apply the given condition. $W_{1} = 0 \Rightarrow cos (3 0^{\circ} + ϕ) = 0 \Rightarrow ϕ = 6 0^{\circ}$ lagging, i.e. power factor $0.5$ lagging. The other root, $ϕ = - 12 0^{\circ}$ , would put $∣ ϕ ∣ > 9 0^{\circ}$ , meaning the load delivers power instead of drawing it. That is excluded.
Reading of $W_{2}$ , with $V_{L} = 400 V$ and $I_{L} = \frac{1}{3} A$ : $W_{2} = 400 \times \frac{1}{3} \times cos (3 0^{\circ} - 6 0^{\circ}) = \frac{400}{3} \times \frac{3}{2} = 200 W$
Check: total power $P = 3 V_{L} I_{L} cos ϕ = 3 \times 400 \times \frac{1}{3} \times 0.5 = 200 W = W_{1} + W_{2}$ , as expected.

KEY IDEA

In the two-wattmeter method, $W_{1} = V_{L} I_{L} cos (3 0^{\circ} + ϕ)$ and $W_{2} = V_{L} I_{L} cos (3 0^{\circ} - ϕ)$ . If one meter reads zero, then $ϕ = 6 0^{\circ}$ (power factor 0.5), and the other meter carries all the power: $W_{2} = V_{L} I_{L} cos 3 0^{\circ} = 3 V_{L} I_{L} cos 6 0^{\circ}$ .

WATCH OUT FOR

One common mistake is using the phase voltage $400/ 3$ in the meter formula instead of the line voltage. Each pressure coil sees a line-to-line voltage, and using phase voltage gives about $115.5 W$ . Another is forgetting that one meter at zero means $ϕ = 6 0^{\circ}$ and the other meter reads the whole power. That leads to wrongly halving the total, or using $cos ϕ = 1$ and getting $400 W$ .

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GATE 2020 IN Q.42: solution and answer

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