An e.m.f. E = E_0 cos omega t is applied to a | vedclass

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(C) The average power consumed in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$. Here,the power factor is $\cos \phi = \frac{R}{Z}$. The

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$\frac{E_0^2}{4 R}$

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(C) The average power consumed in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$. Here,the power factor is $\cos \phi = \frac{R}{Z}$. The $rms$ current is $I_{rms} = \frac{E_{rms}}{Z} = \frac{E_0}{\sqrt{2} Z}$. Substituting these into the power formula: $P = \left( \frac{E_0}{\sqrt{2}} \right) \left( \frac{E_0}{\sqrt{2} Z} \right) \left( \frac{R}{Z} \right) = \frac{E_0^2 R}{2 Z^2} \dots (i)$. Given that the inductive reactance $X_L = R$,the impedance $Z$ of the $L-R$ circuit is: $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2} R$. Substituting $Z^2 = 2 R^2$ into equation $(i)$: $P = \frac{E_0^2 R}{2 (2 R^2)} = \frac{E_0^2 R}{4 R^2} = \frac{E_0^2}{4 R}$.

An e.m.f. $E = E_0 \cos \omega t$ is applied to an $L-R$ circuit. The inductive reactance is equal to the resistance $R$ of the circuit. The power consumed in the circuit is:

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