In a series LR circuit, the inductive reactance is equal to the resistance R of the circuit. An emf \( E = E_0\cos\omega t \) 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) → \( \displaystyle\frac{E_0^{2}}{4R} \)
The average power in an ac circuit is \( P = I_{\mathrm{rms}}^{2}R \), and only the resistor dissipates power. Here \( X_L = R \), so
\( \displaystyle Z = \sqrt{R^{2} + X_L^{2}} = \sqrt{2}\,R \)
\( \displaystyle E_{\mathrm{rms}} = \frac{E_0}{\sqrt{2}} \qquad I_{\mathrm{rms}} = \frac{E_{\mathrm{rms}}}{Z} = \frac{E_0}{\sqrt{2}\times\sqrt{2}\,R} = \frac{E_0}{2R} \)
\( \displaystyle P = I_{\mathrm{rms}}^{2}R = \frac{E_0^{2}}{4R^{2}}\,R = \frac{E_0^{2}}{4R} \)
Check with \( P = E_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi \): here \( \tan\phi = X_L/R = 1 \), so \( \cos\phi = 1/\sqrt{2} \):
\( \displaystyle P = \frac{E_0}{\sqrt{2}} \times \frac{E_0}{2R} \times \frac{1}{\sqrt{2}} = \frac{E_0^{2}}{4R} \)