In an L-R circuit,the inductive reactance is | vedclass

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(C) Given that the inductive reactance $X_L = \sqrt{3} R$.The impedance $Z$ of the $L-R$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$.Substituting $X_

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$\frac{E_0^2}{8 R}$

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(C) Given that the inductive reactance $X_L = \sqrt{3} R$.The impedance $Z$ of the $L-R$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$.Substituting $X_L = \sqrt{3} R$,we get $Z = \sqrt{R^2 + (\sqrt{3} R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.The peak current in the circuit is $I_0 = \frac{E_0}{Z} = \frac{E_0}{2R}$.The average power consumed in an $AC$ circuit is given by $P = I_{rms}^2 R = \left(\frac{I_0}{\sqrt{2}}\right)^2 R$.Substituting $I_0 = \frac{E_0}{2R}$,we get $P = \frac{1}{2} \left(\frac{E_0}{2R}\right)^2 R = \frac{1}{2} \cdot \frac{E_0^2}{4R^2} \cdot R = \frac{E_0^2}{8R}$.

In an $L-R$ circuit,the inductive reactance is equal to $\sqrt{3}$ times the resistance $R$ of the circuit. An e.m.f. $E = E_0 \sin(\omega t)$ is applied to the circuit. The power consumed in the circuit is:

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