Question

An alternating emf of angular frequency $\omega$ is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency

• A. $\frac{\omega}{4}$
• B. $\frac{\omega}{2}$
• C. $2 \omega$
• D. $4 \omega$

Answer 1

Answer: (C)

$ V=V_{0} \sin \omega t$
If it is applied across an inductor, then $I$ lag by $\frac{\pi}{2} .$ Thus
$I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right)$
$P=V I=V_{0} I_{0} \sin \omega t \sin \left(\omega t-\frac{\pi}{2}\right)$
$=-\left(\frac{V_{0} I_{0}}{2}\right) 2 \sin \omega t \cos \omega t$
$P=-\left(\frac{V_{0} I_{0}}{2}\right) \sin (2 \omega t)$
$P=\frac{V_{0} I_{0}}{2} \sin (2 \omega t+\pi)$
$\Rightarrow P$ has an angular frequency of '$2 \omega^{\prime}$'.