SSC JE Electrical Engineering 23 Jan 2018 Evening Paper

Question : 140 of 200

Marks: +1, -0

Power consumed by a balanced star connected 3-phase load is measured using two-wattmeter method. The phase voltage and phese current in the load is 220 V and 10 A respectively. What will be difference in reading (in w) of the two wattmeter, if the power factor of the system is 0.8 lagging?

2286.3
2861.2
3048.4
3810.5

Validate

Solution:

Total power (W)

= 3 \V_\ph \I _\ph \cos \ϕ

.
= 3 × 220 × 10 × .8

W_1 + W_2

= 5280 watt

\tan\ϕ=√3({W_1-W_2}/{W_1+W_2})

\cos\ϕ=0.8,\sin\ϕ=0.6

⇒\tan\ϕ=0.6/0.8=0.75

∴0.75=√3({W_1-W_2}/W)

⇒W_1-W_2={0.75×5280}/1.732

= 2286.3 watt

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