In an ac circuit,V and I are given by V = 100 | vedclass

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(C) The instantaneous voltage is $V = 100 \sin(100t) 	ext{ V}$,so the peak voltage $V_0 = 100 	ext{ V}$.The instantaneous current is $I = 100 \sin(1

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$2.5$

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(C) The instantaneous voltage is $V = 100 \sin(100t) 	ext{ V}$,so the peak voltage $V_0 = 100 	ext{ V}$.The instantaneous current is $I = 100 \sin(100t + \frac{\pi}{3}) 	ext{ mA}$,so the peak current $I_0 = 100 	imes 10^{-3} 	ext{ A} = 0.1 	ext{ A}$.The phase difference between voltage and current is $\phi = \frac{\pi}{3}$.The average power dissipated in an $ac$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.We know that $V_{rms} = \frac{V_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.Substituting the values: $P = \left( \frac{100}{\sqrt{2}} ight) 	imes \left( \frac{0.1}{\sqrt{2}} ight) 	imes \cos\left( \frac{\pi}{3} ight)$.$P = \frac{100 	imes 0.1}{2} 	imes \frac{1}{2} = \frac{10}{4} = 2.5 	ext{ W}$.

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

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