SSC JE Electrical Engineering 26 Sep 2019 Shift 1 Paper

Question : 105 of 200

Marks: +1, -0

If an AC circuit is supplied with an active power of 500 W and reactive power 500 VAR, then the load power factor of the circuit is:

Solution:

P = 500W
Q= 500VAR

\cos φ=\;{P}/{√{P^{2}+Q^{2}}}

=\;{500}/{√{500^{2}+500^{2}}}

=1/√ 2

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