A inductor of reactance `1 Omega` and a resistor of `2 Omega` are connected in series to the terminals of a 6 V (rms) a.c. source. The power dissipated in the circuit is

Options

A. 8 W
B. 12 W
C. `14.4 W`
D. 18 W

Explanation

To solve the problem, we need to calculate the power dissipated in a circuit consisting of a resistor and an inductor connected in series to an AC source. Here are the steps to find the solution:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Reactance of the inductor, \( X_L = 1 \, \Omega \)
- Resistance of the resistor, \( R = 2 \, \Omega \)
- RMS voltage of the AC source, \( V_{rms} = 6 \, V \)

2. **Calculate the Impedance (Z)**:
- The impedance \( Z \) in a series circuit containing a resistor and an inductor is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
- Substituting the values:
\[ Z = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \, \Omega \]

3. **Calculate the RMS Current (\( I_{rms} \))**:
- The RMS current can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
- Substituting the values:
\[ I_{rms} = \frac{6}{\sqrt{5}} \, A \]

4. **Calculate the Power Factor (\( \cos \phi \))**:
- The power factor is given by:
\[ \cos \phi = \frac{R}{Z} \]
- Substituting the values:
\[ \cos \phi = \frac{2}{\sqrt{5}} \]

5. **Calculate the Power Dissipated (P)**:
- The power dissipated in the circuit is given by:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos \phi \]
- Substituting the values:
\[ P = 6 \cdot \left(\frac{6}{\sqrt{5}}\right) \cdot \left(\frac{2}{\sqrt{5}}\right) \]
- Simplifying:
\[ P = 6 \cdot \frac{12}{5} = \frac{72}{5} = 14.4 \, W \]

### Final Answer:
The power dissipated in the circuit is \( 14.4 \, W \).

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