In a $\mathrm{L}-\mathrm{R}$ circuit, the inductive reactance is equal to the resistance ' $R$ ' in the…

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
  1. $\frac{\mathrm{E}_0^2}{\sqrt{2} \mathrm{R}}$
  2. $\frac{E_0^2}{4 R}$
  3. $\frac{\mathrm{E}_0^2}{2 \mathrm{R}}$
  4. $\frac{\mathrm{E}_0^2}{8 \mathrm{R}}$

Solution

$\mathrm{P}=\mathrm{E}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\cos \phi=\frac{R}{Z}$ 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 $X_L=R$ $\therefore \quad \mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{R}^2}$ $=\sqrt{2} \mathrm{R}$ $\therefore \quad P=\frac{E_0^2}{4 R}$

Asked in: MHT CET 2023 (13 May Shift 2)

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