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Q.

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

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a

$\frac{V_{0}^{2}}{2 ω L} \sin ω t$

b

$\frac{V_{0}^{2}}{ω L} \sin 2 ω t$

c

$\frac{- V_{0}^{2}}{2 ω L} \sin 2 ω t$

d

$\frac{- V_{0}^{2}}{2 ω L} \sin ω t$

answer is C.

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Detailed Solution

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$V = L \frac{d I}{d t} \Rightarrow \frac{d I}{d t} = \frac{V_{0}}{L} \sin ω t \Rightarrow I = - \frac{V_{0}}{ω L} \cos ω$ t

Instantaneous power,

$P = V I = - \frac{V_{0}^{2}}{ω L} \sin ω t \cos ω t = - \frac{V_{0}^{2}}{2 ω L} \sin 2 ω t$

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In an AC circuit, voltage V=V0 sin ωt and inductor L is connected across the circuit, Then the instantaneous power will be