An AC voltage V=20 sin 200 pi t is applied to | vedclass

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(D) The given voltage is $V = V_0 \sin(\omega t)$,where $V_0 = 20 \ V$.The given current is $I = I_0 \sin(\omega t + \phi)$,where $I_0 = 10 \ A$ and t

Vedclass · Accepted Answer

$50$

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(D) The given voltage is $V = V_0 \sin(\omega t)$,where $V_0 = 20 \ V$.The given current is $I = I_0 \sin(\omega t + \phi)$,where $I_0 = 10 \ A$ and the phase difference $\phi = \frac{\pi}{3} = 60^{\circ}$.The average power dissipated in an $AC$ circuit is given by the formula $\langle P angle = V_{rms} I_{rms} \cos \phi$.We know that $V_{rms} = \frac{V_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.Substituting the values: $\langle P angle = \frac{20}{\sqrt{2}} 	imes \frac{10}{\sqrt{2}} 	imes \cos(60^{\circ})$.$\langle P angle = \frac{200}{2} 	imes \frac{1}{2} = 100 	imes 0.5 = 50 \ W$.

An $AC$ voltage $V=20 \sin 200 \pi t$ is applied to a series $LCR$ circuit which drives a current $I=10 \sin (200 \pi t + \frac{\pi}{3})$. The average power dissipated is: (in $W$)

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