GATE 2018 EE Q.5: solution and answer

GATE 2018 · EEQ.5 · 1 mark · MCQ

Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeters reads half of the other (both positive), then the power factor of the load is

(A)
0.532
(B)
0.632
(C)
0.707
(D)
0.866

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Answer and solution

ANSWER

The power factor of the load is $cos 3 0^{\circ} = \frac{3}{2} \approx 0.866$ , option (D).

Official answer: DSource: Official answer key, GATE 2018 EE (IIT Guwahati)

Call the larger reading $W_{1} = 2 W$ and the smaller one $W_{2} = W$ . The question says both readings are positive.
For a balanced load measured by the two-wattmeter method, $tan ϕ = \frac{3 ( W _{1} - W _{2} )}{W _{1} + W _{2}}$ , where $W_{1}$ is the larger reading.
Substitute the readings: $tan ϕ = \frac{3 ( 2 W - W )}{2 W + W} = \frac{3}{3} = \frac{1}{3}$
So $ϕ = tan^{- 1} (\frac{1}{3}) = 3 0^{\circ}$ .
Power factor $= cos 3 0^{\circ} = \frac{3}{2} \approx 0.866$ . This is option (D).
Check: the readings are $W_{1} = V_{L} I_{L} cos (3 0^{\circ} - ϕ)$ and $W_{2} = V_{L} I_{L} cos (3 0^{\circ} + ϕ)$ . With $ϕ = 3 0^{\circ}$ these become $V_{L} I_{L} cos 0^{\circ} = V_{L} I_{L}$ and $V_{L} I_{L} cos 6 0^{\circ} = 0.5 V_{L} I_{L}$ . The ratio is 2 : 1 with both readings positive, as the question states. Whether the load is lagging or leading only changes which meter gives the larger reading, so the power factor is 0.866 either way.

KEY IDEA

In the two-wattmeter method on a balanced load, the power factor depends only on the ratio of the two readings: $tan ϕ = \frac{3 ( W _{1} - W _{2} )}{W _{1} + W _{2}}$ . When one reading is half the other and both are positive, $tan ϕ = \frac{1}{3}$ , so $ϕ = 3 0^{\circ}$ and the power factor is 0.866. Points worth remembering: equal readings give a power factor of 1, a ratio of 2 : 1 gives 0.866, one reading at zero gives 0.5, and one negative reading means the power factor is below 0.5.

WATCH OUT FOR

Leaving out the $3$ gives $tan ϕ = \frac{1}{3}$ and a power factor of about 0.949, which is not among the options. Another slip is treating the smaller reading as negative even though the question says both are positive. That gives $tan ϕ = \frac{3 ( 1.5 )}{0.5} = 33$ and a power factor of about 0.189. Some students also put the sum on top and the difference below, which gives $tan ϕ = 33$ as well.

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GATE 2018 EE Q.5: solution and answer

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