A coil of resistance R and inductance L is connected across an a.c power supply of r.m.s. voltage V. The average power dissipated in the coil is,

Options

A. `V^2//R`
B. `(V^2R)/((R^2+omega^2L^2))`
C. `(V^2)/(sqrt(R^2+omega^2L^2))`
D. zero

Answer

To find the average power dissipated in a coil of resistance \( R \) and inductance \( L \) connected to an AC power supply with RMS voltage \( V \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Resistance of the coil: \( R \)
- Inductance of the coil: \( L \)
- RMS Voltage across the coil: \( V \)

2. **Understand the Concept of Impedance**:
- The impedance \( Z \) of the coil in an AC circuit is given by:
\[ Z = \sqrt{R^2 + (\omega L)^2} \]
where \( \omega \) is the angular frequency of the AC supply.

3. **Calculate the RMS Current**:
- The RMS current \( I \) in the circuit can be expressed as:
\[ I = \frac{V}{Z} \]
- Substituting the expression for \( Z \):
\[ I = \frac{V}{\sqrt{R^2 + (\omega L)^2}} \]

4. **Use the Formula for Average Power**:
- The average power \( P_{\text{average}} \) dissipated in the coil can be calculated using the formula:
\[ P_{\text{average}} = I^2 R \]
- Substituting the expression for \( I \):
\[ P_{\text{average}} = \left(\frac{V}{\sqrt{R^2 + (\omega L)^2}}\right)^2 R \]

5. **Simplify the Expression**:
- Squaring the current expression:
\[ P_{\text{average}} = \frac{V^2}{R^2 + (\omega L)^2} R \]
- Thus, the final expression for the average power dissipated in the coil is:
\[ P_{\text{average}} = \frac{V^2 R}{R^2 + (\omega L)^2} \]

### Final Answer:
The average power dissipated in the coil is:
\[ P_{\text{average}} = \frac{V^2 R}{R^2 + (\omega L)^2} \]