If the voltage and current in an ac circuit a | vedclass

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(C) The voltage is given by $V(t) = 50 \sin(50t) 	ext{ V}$,so the peak voltage $V_0 = 50 	ext{ V}$.The current is given by $I(t) = 50 \sin(50t + \fr

Vedclass · Accepted Answer

$0.884$

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(C) The voltage is given by $V(t) = 50 \sin(50t) 	ext{ V}$,so the peak voltage $V_0 = 50 	ext{ V}$.The current is given by $I(t) = 50 \sin(50t + \frac{\pi}{4}) 	ext{ mA}$,so the peak current $I_0 = 50 	ext{ mA} = 50 	imes 10^{-3} 	ext{ A} = 0.05 	ext{ A}$.The phase difference between voltage and current is $\phi = \frac{\pi}{4}$.The average power dissipated in an $AC$ circuit is given by the formula $P = V_{rms} I_{rms} \cos(\phi)$.We know that $V_{rms} = \frac{V_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.Substituting these values,$P = \frac{V_0}{\sqrt{2}} 	imes \frac{I_0}{\sqrt{2}} 	imes \cos(\phi) = \frac{V_0 I_0}{2} \cos(\phi)$.Substituting the given values: $P = \frac{50 	imes 0.05}{2} 	imes \cos(\frac{\pi}{4})$.$P = \frac{2.5}{2} 	imes \frac{1}{\sqrt{2}} = 1.25 	imes 0.707 = 0.88375 	ext{ W}$.Rounding this value,we get $P \approx 0.884 	ext{ W}$.

If the voltage and current in an ac circuit are respectively $50 \sin (50 t) \text{ V}$ and $50 \sin (50 t + \frac{\pi}{4}) \text{ mA}$,then the power dissipated in the circuit is nearly (in $W$)

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