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Power Factor

A lamp consum...

Question

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

A

\frac{π}{6}

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B

\frac{π}{3}

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C

\frac{π}{4}

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D

\frac{π}{2}

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Solution

The correct option is B $\frac{π}{3}$
$P = \frac{1}{2} V_{0} i_{0} c o s ϕ \Rightarrow P = P_{P e a k} . c o s ϕ$
$\Rightarrow \frac{1}{2} (P_{P e a k}) = P_{P e a k} c o s ϕ \Rightarrow c o s ϕ = \frac{1}{2} \Rightarrow ϕ = \frac{π}{3}$

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