Question

A lamp consumers only $50\%$ of maximum power in an A.C. circuit. What is the phase difference between the applied voltage and the circuit current?

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

Answer 1

The correct option is B $\frac{\pi }{3}$

Average power dissipated in lamp $P_{av}=P_{max}\cos \varphi$
where $\varphi$ is the phase difference between the applied voltage and circuit current.
$\therefore$ $0.5P_{max}=P_{max}\cos \varphi$
or $\cos \varphi =0.5$
$⟹\varphi =\frac{\varphi }{3}$
Correct answer is option B.