In an AC circuit,V and I are given by V = 150 | vedclass

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(A) The given equations are $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$.Comparing with the given values,$V_0 = 150 	ext{ V}$,$I_0 =

Vedclass · Accepted Answer

$5625$

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(A) The given equations are $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$.Comparing with the given values,$V_0 = 150 	ext{ V}$,$I_0 = 150 	ext{ A}$,and the phase difference $\phi = \pi/3 = 60^\circ$.The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.Since $V_{rms} = V_0 / \sqrt{2}$ and $I_{rms} = I_0 / \sqrt{2}$,the formula becomes $P = \frac{1}{2} V_0 I_0 \cos \phi$.Substituting the values: $P = \frac{1}{2} 	imes 150 	imes 150 	imes \cos(60^\circ)$.Since $\cos(60^\circ) = 0.5$,we get $P = 0.5 	imes 150 	imes 150 	imes 0.5$.$P = 0.25 	imes 22500 = 5625 	ext{ W}$.

In an $AC$ circuit,$V$ and $I$ are given by $V = 150 \sin(150t) \text{ V}$ and $I = 150 \sin(150t + \pi/3) \text{ A}$. The power dissipated in the circuit is: (in $\text{ W}$)

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