In a series LR circuit, the inductive reactance is equal to the resistance R of the circuit. An emf $E = E_0\cos wt$ is applied to the circuit. The average power dissipated in one cycle in the circuit is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{{E_0}^2}{4R}$
Given: In a series $LR$ circuit, $X_L = R$.
Applied emf: $E = E_0 \cos \omega t$
Impedance: $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2}R$
RMS voltage: $V_{rms} = \frac{E_0}{\sqrt{2}}$
RMS current: $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_0/\sqrt{2}}{\sqrt{2}R} = \frac{E_0}{2R}$
Average power dissipated: $P = I_{rms}^2 R = \left(\frac{E_0}{2R}\right)^2 R = \frac{E_0^2}{4R}$
Final Answer: $\;\; P = \frac{E_0^2}{4R}$