When an inductor is connected to an AC source | vedclass

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Question

(D) The voltage across the inductor is $E = E_0 \sin(\omega t)$.In a pure inductor,the current lags the voltage by $\pi/2$,so $i = i_0 \sin(\omega t -

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$2\omega$

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(D) The voltage across the inductor is $E = E_0 \sin(\omega t)$.In a pure inductor,the current lags the voltage by $\pi/2$,so $i = i_0 \sin(\omega t - \pi/2) = -i_0 \cos(\omega t)$.The instantaneous power $P$ is given by $P = E \cdot i$.$P = (E_0 \sin(\omega t)) \cdot (-i_0 \cos(\omega t))$.$P = -E_0 i_0 \sin(\omega t) \cos(\omega t)$.Using the trigonometric identity $\sin(2	heta) = 2 \sin(	heta) \cos(	heta)$,we get:$P = -\frac{E_0 i_0}{2} \sin(2\omega t)$.The frequency of the instantaneous power is the coefficient of $t$ divided by $2\pi$. Since the angular frequency of the power is $2\omega$,the frequency is $2\omega$.

When an inductor is connected to an $AC$ source of frequency $\omega$,what is the frequency of the instantaneous power?

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