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In a $\mathrm{L}-\mathrm{R}$ circuit, the inductive reactance is equal to the resistance ' $R$ ' in the circuit. An emf $\mathrm{E}=\mathrm{E}_0 \cos \omega \mathrm{t}$ is applied to the circuit. The power consumed in the circuit is
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The correct answer is:
$\frac{E_0^2}{4 R}$
$\mathrm{P}=\mathrm{E}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$
$\cos \phi=\frac{R}{Z}$
Also, $I_{\text {rms }}=\frac{E_{\text {rms }}}{Z}=\frac{E_0}{Z \sqrt{2}}$
$\therefore \quad P=\frac{E_0}{\sqrt{2}} \times \frac{E_0}{Z \sqrt{2}} \times \frac{R}{Z}$
$=\frac{E_0^2 R}{2 Z^2}$
Given $X_L=R$
$\therefore \quad \mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{R}^2}$
$=\sqrt{2} \mathrm{R}$
$\therefore \quad P=\frac{E_0^2}{4 R}$
$\cos \phi=\frac{R}{Z}$
Also, $I_{\text {rms }}=\frac{E_{\text {rms }}}{Z}=\frac{E_0}{Z \sqrt{2}}$
$\therefore \quad P=\frac{E_0}{\sqrt{2}} \times \frac{E_0}{Z \sqrt{2}} \times \frac{R}{Z}$
$=\frac{E_0^2 R}{2 Z^2}$
Given $X_L=R$
$\therefore \quad \mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{R}^2}$
$=\sqrt{2} \mathrm{R}$
$\therefore \quad P=\frac{E_0^2}{4 R}$
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