SSC JE Electrical Engineering 26 Sep 2019 Shift 1 Paper

Question : 158 of 200

Marks: +1, -0

In the two wattmeter method, the readings of the two wattmeter are 500W, 500W respectively.
The load power factor in a balanced 3-phase 3- wire circuit is:

Validate

Solution:

\tan ϕ = √3({W_1-W_2}/{W_1+W_2})

\tan ϕ = √3({500-500}/{500+500})

∴

\tan ϕ=0

∴

ϕ=0°

∴ Power factor cosϕ = cos 0 = 1

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