An e.m.f. $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ is applied to a circuit containing 'L' and 'R'…

An e.m.f. $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ is applied to a circuit containing 'L' and 'R' in series. If $X_{L}=R$, then the power dissipated in the circuit is
  1. $\frac{\mathrm{E}_{0}^{2}}{4 \mathrm{R}}$
  2. $\frac{\mathrm{E}_{0}}{2 \mathrm{R}}$
  3. $\frac{\mathrm{E}_{0}}{4 \mathrm{R}}$
  4. $\frac{\mathrm{E}_{0}^{2}}{2 \mathrm{R}}$

Solution

$\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ $P=\frac{E_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \cos \theta$ $=\frac{E_{0}}{\sqrt{2}} \times \frac{E_{0}}{\sqrt{2} Z} \times \frac{R}{Z}$ $=\frac{E_{0}^{2} R}{2 Z^{2}} \quad \quad Z^{2}=R^{2}+\omega L^{2}=2 R^{2}$ $=\frac{E_{0}^{2} R}{4 R^{2}}=\frac{E_{0}^{2}}{4 R}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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