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 find the power dissipated in the circuit with an inductor and a resistor connected in series to an AC source, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Inductive reactance, \( X_L = 1 \, \Omega \)
- Resistance, \( R = 2 \, \Omega \)
- RMS voltage, \( V_{rms} = 6 \, V \)

2. **Calculate Impedance (Z):**
The impedance \( Z \) of 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 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 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 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:
\[ P = 6 \cdot \frac{36}{5} = \frac{216}{5} = 43.2 \, W \]

6. **Final Calculation:**
The final value of power dissipated is:
\[ P = 14.4 \, W \]

### Final Answer:
The power dissipated in the circuit is **14.4 Watts**.