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

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(C) The given equations are $V = V_m \sin(\omega t)$ and $I = I_m \sin(\omega t + \phi)$.Comparing with the given equations,$V_m = 150 \, V$,$I_m = 15

Vedclass · Accepted Answer

$5625$

Vedclass · Answer

(C) The given equations are $V = V_m \sin(\omega t)$ and $I = I_m \sin(\omega t + \phi)$.Comparing with the given equations,$V_m = 150 \, V$,$I_m = 150 \, A$,and the phase difference $\phi = \frac{\pi}{3}$.The $RMS$ values are $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{150}{\sqrt{2}} \, V$ and $I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{150}{\sqrt{2}} \, A$.The power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.Substituting the values: $P = \left( \frac{150}{\sqrt{2}} ight) 	imes \left( \frac{150}{\sqrt{2}} ight) 	imes \cos\left( \frac{\pi}{3} ight)$.Since $\cos(60^{\circ}) = 0.5$,we have $P = \frac{150 	imes 150}{2} 	imes 0.5 = \frac{22500}{4} = 5625 \, W$.

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

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