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In a three phase power measurement by two waltmeter method , both the wattmeters read the same value. What is the power factor of the load? Justify your answer.
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When both watt meters show same value, P.f =1

Proof:

$\begin{align*} P.f. &= \cos \left [ \tan^{-1} \left ( \sqrt{3}.\dfrac{W_1-W_2}{W_1+W_2} \right ) \right ]\\ &=\cos \left [ \tan ^{-1}\left ( \dfrac{0}{W_1+W_2} \right ) \right ]\\ &=\cos (\tan ^{-1}0)\\ &=\cos 0 =1 \end{align*}$

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