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))`

Answer

To solve the problem of finding the power dissipated in a series circuit consisting of an inductor and a resistor connected to an AC source, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components:**
- We have an inductor with inductance \(L\) and a resistor with resistance \(R\) connected in series to an AC source with frequency \(\omega\).

2. **Calculate the Inductive Reactance:**
- The inductive reactance \(X_L\) is given by the formula:
\[ X_L = \omega L \]

3. **Calculate the Impedance \(Z\):**
- The total impedance \(Z\) of the series circuit is calculated using the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
- Substituting \(X_L\) into the equation, we get:
\[ Z = \sqrt{R^2 + (\omega L)^2} \]

4. **Determine the Power Dissipated:**
- The power \(P\) dissipated in the circuit can be expressed as:
\[ P = V_{\text{rms}} I_{\text{rms}} \cos \phi \]
- Here, \(V_{\text{rms}}\) is the root mean square voltage, \(I_{\text{rms}}\) is the root mean square current, and \(\cos \phi\) is the power factor.

5. **Express \(I_{\text{rms}}\) and \(\cos \phi\):**
- The current \(I_{\text{rms}}\) can be expressed in terms of the voltage and impedance:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]
- The power factor \(\cos \phi\) is given by:
\[ \cos \phi = \frac{R}{Z} \]

6. **Substituting into the Power Formula:**
- Substituting \(I_{\text{rms}}\) and \(\cos \phi\) into the power formula gives:
\[ P = V_{\text{rms}} \left(\frac{V_{\text{rms}}}{Z}\right) \left(\frac{R}{Z}\right) \]
- Simplifying this, we have:
\[ P = \frac{V_{\text{rms}}^2 R}{Z^2} \]

7. **Substituting the Expression for \(Z\):**
- Now, substituting the expression for \(Z\) into the power equation:
\[ P = \frac{V_{\text{rms}}^2 R}{R^2 + (\omega L)^2} \]

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