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Power Factor

A lamp consum...

Question

A lamp consumes only $50 %$ of peak power in an ac circuit. What is the phase difference between the applied voltage and the circuit current?

\frac{π}{6}

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\frac{π}{3}

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\frac{π}{4}

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\frac{π}{2}

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Solution

The correct option is A $\frac{π}{3}$
$P = \frac{1}{2} v_{0} i_{0} c o s ψ$ ,

$P = P_{p e a k} . c o s ψ$

$\frac{1}{2} P_{p e a k} = P_{p e a k} c o s ψ = \frac{1}{2} = ψ = \frac{π}{3}$

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\frac{π}{3}

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\frac{π}{3}

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