In an A.C. circuit,V and I are given by V = 1 | vedclass

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(C) The given equations are $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$.Here,$V_0 = 100 \ V$,$I_0 = 100 \ mA = 0.1 \ A$,and the phase

Vedclass · Accepted Answer

$2.5$

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(C) The given equations are $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$.Here,$V_0 = 100 \ V$,$I_0 = 100 \ mA = 0.1 \ A$,and the phase difference $\phi = \frac{\pi}{3}$.The average power dissipated in an $A.C.$ circuit is given by $P = V_{rms} 	imes I_{rms} 	imes \cos(\phi)$.$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \ V$.$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} \ A$.$P = \left(\frac{100}{\sqrt{2}}ight) 	imes \left(\frac{0.1}{\sqrt{2}}ight) 	imes \cos\left(\frac{\pi}{3}ight)$.$P = \frac{100 	imes 0.1}{2} 	imes \frac{1}{2} = \frac{10}{4} = 2.5 \ W$.

In an $A.C.$ circuit,$V$ and $I$ are given by $V = 100 \sin(100t) \ V$ and $I = 100 \sin(100t + \frac{\pi}{3}) \ mA$. The power dissipated in the circuit is ...... $W$.

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