When a voltage V_{s} = 200 sqrt{2} sin(omega | vedclass

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Question

(C) The average power consumed in an $AC$ circuit is given by the formula $P_{av} = V_{rms} I_{rms} \cos \phi$.Given voltage $V_{s} = 200 \sqrt{2} \si

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$100 \sqrt{6} 	ext{ watt}$

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(C) The average power consumed in an $AC$ circuit is given by the formula $P_{av} = V_{rms} I_{rms} \cos \phi$.Given voltage $V_{s} = 200 \sqrt{2} \sin(\omega t + 15^{\circ})$,the peak voltage $V_{0} = 200 \sqrt{2} 	ext{ V}$.Thus,$V_{rms} = \frac{V_{0}}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 	ext{ V}$.Given current $I = 2 \sin(\omega t + 45^{\circ})$,the peak current $I_{0} = 2 	ext{ A}$.Thus,$I_{rms} = \frac{I_{0}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} 	ext{ A}$.The phase difference $\phi$ is the difference between the phase of voltage and current: $\phi = 45^{\circ} - 15^{\circ} = 30^{\circ}$.Substituting these values into the power formula:$P_{av} = 200 	imes \sqrt{2} 	imes \cos(30^{\circ})$$P_{av} = 200 	imes \sqrt{2} 	imes \frac{\sqrt{3}}{2} = 100 \sqrt{6} 	ext{ watt}$.

When a voltage $V_{s} = 200 \sqrt{2} \sin(\omega t + 15^{\circ})$ is applied to an $AC$ circuit,the current in the circuit is found to be $I = 2 \sin(\omega t + 45^{\circ})$. The average power consumed in the circuit is:

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