Question

In an $L - R$ circuit, the inductive reactance is equal to $\sqrt{3}$ times the resistance '$R$' of the circuit. An e.m.f. $\text{E} = \text{E}_0 \sin(\omega t)$ is applied to the circuit. The power consumed in the circuit is

• A. $\frac{\text{E}_0^2}{4\text{R}}$
• B. $\frac{E_0^2}{6R}$
• C. $\frac{\text{E}_0^2}{8\text{R}}$
• D. $\frac{E_0^2}{12R}$

Hint:

Power is only dissipated in the resistance $R$, never in the pure inductor $L$.

Answer 1

The Correct Option is C

Step 1: Concept
Average power $P = V_{rms} I_{rms} \cos \phi = \frac{E_0}{\sqrt{2}} \cdot \frac{E_0}{\sqrt{2}Z} \cdot \frac{R}{Z} = \frac{E_0^2 R}{2Z^2}$.

Step 2: Meaning
Impedance $Z = \sqrt{R^2 + X_L^2}$. Given $X_L = \sqrt{3}R$.

Step 3: Analysis
$Z^2 = R^2 + (\sqrt{3}R)^2 = 4R^2$.
$P = \frac{E_0^2 R}{2(4R^2)} = \frac{E_0^2 R}{8R^2}$.

Step 4: Conclusion
The power consumed is $\frac{E_0^2}{8R}$. Final Answer: (C)