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

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(C) The average power in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$.Here,$E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\

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

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(C) The average power in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$.Here,$E_{rms} = \frac{E_0}{\sqrt{2}}$ 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,$P = \frac{E_0}{\sqrt{2}} 	imes \frac{E_0}{Z\sqrt{2}} 	imes \frac{R}{Z} = \frac{E_0^2 R}{2Z^2}$.Given that the inductive reactance $X_L = R$,the impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2}R$.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 $R$. If the applied voltage is $E = E_0 \cos(\omega t)$,what is the power dissipated in the circuit?

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