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

Answer 1

Answer: (C)

If $\frac{dI}{dt}$ is the rate of change of current through inductor L of any instant, then induced emf in the inductor of the same instant is
$=\left(-\frac{LdI}{dt}\right)$
To maintain the flow of current in the circuit applied voltage must be equal and opposite to the induced voltage.
$\therefore V=-\left(-L\frac{dI}{dt}\right).$
Given, $V=V_{0}sin\omega t...\left(i\right)$
$\therefore V_{0}sin\omega t=\frac{LdI}{dt}$
$\frac{dI}{dt}=\frac{V_0}{L}sin\omega t$
or $I=\frac{-V_0}{L}\frac{cos\omega t}{\omega }...\left(ii\right)$
Instantaneous power,
$\because P=VI$
$=\left(V_{0}sin\omega t\right)\left(-\frac{V_{0}cos\omega t}{L\omega }\right)\left(Using\left(i\right)\&\left(ii\right)\right)$
$=\frac{-V_0^{2}sin2\omega t}{2L\omega }$