Question

In an AC circuit, voltage $V=V_{0}sin\omega t$ and inductor L is connected across the circuit.Then the instantaneous power will be

• A. $\frac{V_0^2}{2\omega L}sin\omega t$
• B. $\frac{-V_{-}^2}{2\omega L}sin\omega t$
• C. $\frac{-V_0^2}{2\omega L}sin2\omega t$
• D. $\frac{V_0^2}{\omega L}sin2\omega t$

Answer 1

The correct option is C $\frac{-V_0^2}{2\omega L}sin2\omega t$

$V=L\frac{dI}{dt}$

$\Rightarrow \frac{dI}{dt}=\frac{V_0}{L}sin\omega t$

$\Rightarrow I=-\frac{V_0}{\omega L}cos\omega t$

Instantaneous power,

$P=VI=--\frac{V_0^2}{\omega L}sin\omega tcos\omega t$

$=-\frac{V_0^2}{2\omega L}sin2\omega t$