Question

An e.m.f. $E = E_0 \cos \omega t$ is applied to circuit containing L and R in series. If $X_L = 2R$, then the power dissipated in the circuit is

• A. $\frac{E_0^2}{12R}$
• B. $\frac{E_0^2}{10R}$
• C. $\frac{E_0^2}{8R}$
• D. $\frac{E_0^2}{6R}$

Answer 1

Answer: (B)

$\frac{E_0^2}{10R}$

Answer 2

Solution:

1. Calculate the Impedance:

   $$Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = R\sqrt{5}.$$

2. Determine the Peak Current:

   $$I_0 = \frac{E_0}{Z} = \frac{E_0}{R\sqrt{5}}.$$

3. Calculate the Average Power Dissipated:
   Only the resistor dissipates power; the average power is given by

   $$P_{\text{avg}} = \frac{1}{2} I_0^2 R = \frac{1}{2} \left(\frac{E_0}{R\sqrt{5}}\right)^2 R = \frac{1}{2} \frac{E_0^2}{5R^2} R = \frac{E_0^2}{10R}.$$

Core Explanation:

• Total impedance: $Z = R\sqrt{5}$.
• Peak current: $I_0 = \frac{E_0}{R\sqrt{5}}$.
• Average power: $P_{\text{avg}} = \frac{1}{2} I_0^2 R = \frac{E_0^2}{10R}$.