In an AC circuit,the voltage and current are | vedclass

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(A) The given equations are $V = V_0 \sin(\omega t + \phi_1)$ and $i = i_0 \sin(\omega t + \phi_2)$.Here,$V_0 = 200 	ext{ V}$,$i_0 = 50 	ext{ mA} =

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$2.5$

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(A) The given equations are $V = V_0 \sin(\omega t + \phi_1)$ and $i = i_0 \sin(\omega t + \phi_2)$.Here,$V_0 = 200 	ext{ V}$,$i_0 = 50 	ext{ mA} = 50 	imes 10^{-3} 	ext{ A} = 0.05 	ext{ A}$.The phase difference $\phi = \phi_1 - \phi_2 = -\frac{\pi}{6} - \frac{\pi}{6} = -\frac{\pi}{3}$.The average power dissipated is given by $P = V_{rms} I_{rms} \cos \phi = \frac{V_0}{\sqrt{2}} \frac{i_0}{\sqrt{2}} \cos \phi = \frac{V_0 i_0}{2} \cos \phi$.Substituting the values: $P = \frac{200 	imes 0.05}{2} \cos(-\frac{\pi}{3}) = \frac{10}{2} 	imes \frac{1}{2} = 5 	imes 0.5 = 2.5 	ext{ W}$.

In an $AC$ circuit,the voltage and current are described by $V = 200 \sin(314t - \frac{\pi}{6}) \text{ V}$ and $i = 50 \sin(314t + \frac{\pi}{6}) \text{ mA}$ respectively. The average power dissipated in the circuit is $...... \text{ W}$.

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