In a balanced three phase circuit, power is measured by two wattmeters, the ratio of two wattmeter readings is 2:1. Determine the power factor of the system.

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written 2.9 years ago by

teamques10 ★ 64k

modified 2.4 years ago by

pedsangini276 • 4.7k

basic electrical engineering

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1 Answer

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written 2.9 years ago by

teamques10 ★ 64k

Given:

$\dfrac{W_1}{W_2} =\dfrac{2}1 \ \ \ i.e W_1=2W_2$

To find: Power factor

$\phi{}= tan^{-1} \Big[ \dfrac{\sqrt{3}\left(W_1-W_2\right)}{W_1+W_2} \Big] = {tan}^{-1}\Big[ \dfrac{\sqrt{3}\left(W_1-W_2\right)}{W_1+W_2}\Big] = tan^{-1} \dfrac{\sqrt{3}W_2 }{3W_2} \\[2ex] ϕ=tan^{-1}⁡(0.557)=30^{\circ} \\[2ex] Power \ factor=cos⁡ϕ=cos30^{\circ}=0.8661$

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