An inductor of inductance `L` and ressistor of resistance `R` are joined in series and connected by a source of frequency `omega`. Power dissipated in the circuit is

Options

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

Explanation

To find the power dissipated in a circuit consisting of an inductor of inductance \( L \) and a resistor of resistance \( R \) connected in series to an AC source of frequency \( \omega \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Power Formula in AC Circuits**:
The power \( P \) dissipated in an AC circuit can be expressed as:
\[ P = V_{\text{rms}} I_{\text{rms}} \cos \phi \]
where:
- \( V_{\text{rms}} \) is the root mean square voltage,
- \( I_{\text{rms}} \) is the root mean square current,
- \( \phi \) is the phase difference between the voltage and the current.

2. **Relate Current and Voltage**:
The current \( I \) can be expressed in terms of the voltage \( V \) and the impedance \( Z \) of the circuit:
\[ I = \frac{V}{Z} \]
Therefore, we can rewrite the power equation as:
\[ P = V_{\text{rms}} \left(\frac{V_{\text{rms}}}{Z}\right) \cos \phi = \frac{V_{\text{rms}}^2}{Z} \cos \phi \]

3. **Determine the Impedance \( Z \)**:
For a series RL circuit, the impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( X_L \) is the inductive reactance, defined as:
\[ X_L = \omega L \]
Thus, we can write:
\[ Z = \sqrt{R^2 + (\omega L)^2} \]

4. **Calculate the Power Factor**:
The power factor \( \cos \phi \) for an RL circuit is given by:
\[ \cos \phi = \frac{R}{Z} \]

5. **Substitute into the Power Formula**:
Now substituting \( Z \) and \( \cos \phi \) into the power equation:
\[ P = \frac{V_{\text{rms}}^2}{Z} \cdot \frac{R}{Z} \]
This simplifies to:
\[ P = \frac{V_{\text{rms}}^2 R}{Z^2} \]

6. **Substitute for \( Z^2 \)**:
Substitute \( Z^2 \) into the equation:
\[ Z^2 = R^2 + (\omega L)^2 \]
Therefore, the power dissipated in the circuit becomes:
\[ P = \frac{V_{\text{rms}}^2 R}{R^2 + (\omega L)^2} \]

### Final Result:
The power dissipated in the circuit is:
\[ P = \frac{V_{\text{rms}}^2 R}{R^2 + (\omega L)^2} \]