GATE 2018 IN Q.21: solution and answer

GATE 2018 · INQ.21 · 1 mark · NAT

A 300 V, 5A, 0.2 pf low power factor wattmeter is used to measure the power consumed by a load. The wattmeter scale has 150 divisions and the pointer is on the 100 $^{th}$ division. The power consumed by the load (in Watts) is ______ .

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Answer and solution

ANSWER

The power consumed by the load is 200 W.

Official answer: 200Source: Official answer key, GATE 2018 IN (IIT Guwahati)

A low power factor (LPF) wattmeter is calibrated so that full-scale deflection corresponds to rated voltage × rated current × rated power factor, not to the product $V I$ alone.
Full-scale power: $P_{f s} = V_{r a t e d} I_{r a t e d} cos ϕ_{r a t e d} = 300 \times 5 \times 0.2 = 300 W$
The scale has 150 divisions, so the multiplying factor (watts per division) is $k = \frac{300 W}{150 div} = 2 W/div$
Wattmeter deflection is proportional to power, so the reading at the 100th division is $P = 100 div \times 2 W/div = 200 W$
Power consumed by the load $= 200 W$ .

KEY IDEA

For an LPF wattmeter, full-scale reading = rated $V \times$ rated $I \times$ rated power factor. Here that is $300 \times 5 \times 0.2 = 300 W$ across 150 divisions, so each division is worth $2 W$ .

WATCH OUT FOR

Leaving out the rated power factor and taking full scale as $V I = 1500 W$ . That gives $10 W/div$ and a wrong answer of $1000 W$ . A second slip is to multiply the reading by 0.2 again. The rated power factor only sets the full-scale value; it is not applied to the reading.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 8 V, I, power & energy questions, with notes

GATE 2018 IN Q.21: solution and answer

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