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

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(B) The impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$.Given $X_L = 2R$,we have $Z^2 = R^2 + (2R)^2 = R^2 + 4R^2 = 5R^2$

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$\frac{E_0^2}{10 R}$

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(B) The impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$.Given $X_L = 2R$,we have $Z^2 = R^2 + (2R)^2 = R^2 + 4R^2 = 5R^2$.The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi = \frac{R}{Z}$ is the power factor.Since $V_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_0}{\sqrt{2}Z}$,the power is $P = \frac{E_0}{\sqrt{2}} 	imes \frac{E_0}{\sqrt{2}Z} 	imes \frac{R}{Z} = \frac{E_0^2 R}{2 Z^2}$.Substituting $Z^2 = 5R^2$,we get $P = \frac{E_0^2 R}{2(5R^2)} = \frac{E_0^2 R}{10 R^2} = \frac{E_0^2}{10 R}$.

An e.m.f. $E=E_0 \cos \omega t$ is applied to a circuit containing $L$ and $R$ in series. If $X_L=2 R$,then the power dissipated in the circuit is

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