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 1: Understand the Circuit
The circuit consists of a resistor \( R \) and an inductor \( L \) connected in series to an AC voltage source. The total impedance \( Z \) of the circuit is given by:
\[ Z = \sqrt{R^2 + (\omega L)^2} \]
where \( \omega \) is the angular frequency of the AC supply.

### Step 2: Calculate the RMS Current
The RMS current \( I_{\text{rms}} \) flowing through the circuit can be calculated using Ohm's law:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{V}{\sqrt{R^2 + (\omega L)^2}} \]

### Step 3: Calculate the Average Power
The average power \( P \) dissipated in the resistor can be calculated using the formula:
\[ P = I_{\text{rms}}^2 \cdot R \]
Substituting \( I_{\text{rms}} \) from the previous step:
\[ P = \left(\frac{V}{\sqrt{R^2 + (\omega L)^2}}\right)^2 \cdot R \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ P = \frac{V^2}{R^2 + (\omega L)^2} \cdot R \]
Thus, the average power dissipated in the coil is:
\[ P = \frac{V^2 R}{R^2 + (\omega L)^2} \]

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

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