In an A C circuit, voltage V=V_0 ω t and inductor L is connected...

In an

A C

circuit, voltage

V=V_0 \sin \omega t

and inductor

L

is connected across the circuit. Then the instantaneous power will be In an A C circuit, voltage V=V_0 ω t and inductor L is connected across the circuit. Then the instantaneous power will

In an A C circuit, voltage V=V_0 ω t and inductor L is connected across the circuit. Then the instantaneous power will

Options

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 $\frac{V_0^2}{\omega L} \sin 2 \omega t$

Correct answer

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

Step-by-step solution

\frac{d I}{d t} is the rate of change of current through inductor

of any instant, then induced emf in the inductor of the same instant is =\left(-\frac{L d I}{d t}\right)

To maintain the flow of current in the circuit applied voltage must be equal and opposite to the induced voltage.

\therefore \quad V=-\left(-L \frac{d I}{d t}\right) \Subset

Given,

V=V_0 \sin \omega t

\therefore \quad V_0 \sin \omega t=\frac{L d I}{d t}

\frac{d I}{d t}=\frac{V_0}{L} \sin \omega t

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

Instantaneous power,

\because \quad P=V I

=\left(V_0 \sin \omega t\right)\left(-\frac{V_0 \cos \omega t}{L \omega}\right) \quad(\mathrm{Using}

(i) \& (ii))

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

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