In an LR-circuit,the inductive reactance is e | vedclass

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(C) The average power consumed in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.We know that $E_{r

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

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(C) The average power consumed in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.We know that $E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{E_0}{Z\sqrt{2}}$.The power factor is $\cos \phi = \frac{R}{Z}$.Substituting these values,we get $P = \left(\frac{E_0}{\sqrt{2}}\right) \left(\frac{E_0}{Z\sqrt{2}}\right) \left(\frac{R}{Z}\right) = \frac{E_0^2 R}{2Z^2}$.Given that the inductive reactance $X_L = R$,the impedance $Z$ is calculated as $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2}R$.Substituting $Z^2 = 2R^2$ into the power formula: $P = \frac{E_0^2 R}{2(2R^2)} = \frac{E_0^2 R}{4R^2} = \frac{E_0^2}{4R}$.

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

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