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: Understanding the Concept:
In an AC circuit containing a resistor and an inductor, power is dissipated only by the resistor.
The average power consumed depends on the RMS voltage, RMS current, and the power factor of the circuit.
Step 2: Key Formula or Approach:
Impedance of the circuit: $Z = \sqrt{R^2 + X_L^2}$.
RMS Voltage: $E_{\text{rms}} = \frac{E_0}{\sqrt{2}}$.
Average power: $P = E_{\text{rms}} I_{\text{rms}} \cos\phi = \frac{E_{\text{rms}}^2}{Z} \left(\frac{R}{Z}\right) = \frac{E_{\text{rms}}^2}{Z^2} R$.
Step 3: Detailed Explanation:
We are given that inductive reactance $X_L = \sqrt{3}R$.
First, let's calculate the total impedance $Z$ of the circuit: \[ Z = \sqrt{R^2 + X_L^2} \] Substitute $X_L = \sqrt{3}R$: \[ Z = \sqrt{R^2 + (\sqrt{3}R)^2} \] \[ Z = \sqrt{R^2 + 3R^2} \] \[ Z = \sqrt{4R^2} = 2R \] The RMS voltage of the applied e.m.f is: \[ E_{\text{rms}} = \frac{E_0}{\sqrt{2}} \] The formula for the average power consumed in the circuit is: \[ P = \frac{E_{\text{rms}}^2}{Z^2} R \] Substitute the expressions for $E_{\text{rms}}$ and $Z$: \[ P = \frac{(E_0/\sqrt{2})^2}{(2R)^2} R \] Expand the squared terms: \[ P = \frac{E_0^2 / 2}{4R^2} R \] Simplify the expression: \[ P = \frac{E_0^2}{2 \cdot 4R^2} R \] \[ P = \frac{E_0^2}{8R^2} R \] Cancel one $R$ from the numerator and denominator: \[ P = \frac{E_0^2}{8R} \] Step 4: Final Answer:
The power consumed in the circuit is $\frac{\text{E}_0^2}{8\text{R}}$.