In an AC circuit,V and I are given below. Fin | vedclass

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(A) The power dissipated in an $AC$ circuit is given by the formula: $P = V_{rms} I_{rms} \cos \phi$ Given: $V_0 = 50 \ V$ $I_0 = 50 \ mA = 50 	imes

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$0.625$

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(A) The power dissipated in an $AC$ circuit is given by the formula: $P = V_{rms} I_{rms} \cos \phi$ Given: $V_0 = 50 \ V$ $I_0 = 50 \ mA = 50 	imes 10^{-3} \ A$ Phase difference $\phi = \frac{\pi}{3}$ The $RMS$ values are $V_{rms} = \frac{V_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$. Substituting these into the power formula: $P = \frac{V_0}{\sqrt{2}} 	imes \frac{I_0}{\sqrt{2}} 	imes \cos \phi = \frac{V_0 I_0}{2} \cos \phi$ $P = \frac{50 	imes 50 	imes 10^{-3}}{2} 	imes \cos(\frac{\pi}{3})$ Since $\cos(\frac{\pi}{3}) = 0.5$: $P = \frac{2500 	imes 10^{-3}}{2} 	imes 0.5 = 1.25 	imes 0.5 = 0.625 \ W$

In an $AC$ circuit,$V$ and $I$ are given below. Find the power dissipated in the circuit:
$V = 50 \sin(50t) \ V$
$I = 50 \sin(50t + \frac{\pi}{3}) \ mA$ (in $W$)

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In an $AC$ circuit,$V$ and $I$ are given below. Find the power dissipated in the circuit: $V = 50 \sin(50t) \ V$ $I = 50 \sin(50t + \frac{\pi}{3}) \ mA$ (in $W$)