Question

A lamp consumes only $50$% of maximum power applied in an AC circuit. What will be the phase difference between applied voltage and circuit current?

• A. $\frac{\pi }{3}rad$
• B. $\frac{\pi }{6}rad$
• C. $\frac{\pi }{4}rad$
• D. $\frac{\pi }{2}rad$

Answer 1

The correct option is A $\frac{\pi }{3}rad$

We know that
$\cos \varphi =\eta$
($\because \eta =50$% where, $\varphi$ is the phase difference)
$\cos \varphi =50$%
$\cos \varphi =\frac{1}{2}$
$\cos \varphi =\cos {60}^o\Rightarrow \theta =\frac{\pi }{3}$