An inductor and a resistor of 25 Omega are co | vedclass

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(C) The given voltage is $V(t) = 100 \sin(100 \pi t) \ V$. The peak voltage is $V_0 = 100 \ V$.The root mean square voltage is $V_{rms} = \frac{V_0}{\

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$50$

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(C) The given voltage is $V(t) = 100 \sin(100 \pi t) \ V$. The peak voltage is $V_0 = 100 \ V$.The root mean square voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \ V$.The impedance of the circuit is $Z = 50 \ \Omega$ and the resistance is $R = 25 \ \Omega$.The root mean square current in the circuit is $I_{rms} = \frac{V_{rms}}{Z} = \frac{100/\sqrt{2}}{50} = \frac{2}{\sqrt{2}} = \sqrt{2} \ A$.The average power dissipated in an $AC$ circuit is given by $P = I_{rms}^2 R$.Substituting the values,$P = (\sqrt{2})^2 	imes 25 = 2 	imes 25 = 50 \ W$.

An inductor and a resistor of $25 \Omega$ are connected in series to an ac source of voltage $100 \sin (100 \pi t) \ V$. If the impedance of the circuit is $50 \Omega$,the average power dissipated per cycle in the circuit is: (in $W$)

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