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(A) The instantaneous e.m.f. is given by $E = E_{0} \sin \omega t$.The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \c

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$\frac{E_{0}^{2}}{4 R}$

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(A) The instantaneous e.m.f. is given by $E = E_{0} \sin \omega t$.The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.Here,$V_{rms} = \frac{E_{0}}{\sqrt{2}}$ and $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_{0}}{\sqrt{2} Z}$.The power factor is $\cos \phi = \frac{R}{Z}$.Thus,$P = \left( \frac{E_{0}}{\sqrt{2}} \right) \left( \frac{E_{0}}{\sqrt{2} Z} \right) \left( \frac{R}{Z} \right) = \frac{E_{0}^{2} R}{2 Z^{2}}$.For an $LR$ series circuit,the impedance $Z$ is given by $Z = \sqrt{R^{2} + X_{L}^{2}}$.Given $X_{L} = R$,we have $Z = \sqrt{R^{2} + R^{2}} = \sqrt{2 R^{2}} = R \sqrt{2}$.Therefore,$Z^{2} = 2 R^{2}$.Substituting $Z^{2}$ into the power equation: $P = \frac{E_{0}^{2} R}{2 (2 R^{2})} = \frac{E_{0}^{2} R}{4 R^{2}} = \frac{E_{0}^{2}}{4 R}$.

An e.m.f. $E = E_{0} \sin \omega t$ is applied to a circuit containing $L$ and $R$ in series. If $X_{L} = R$,then the power dissipated in the circuit is

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