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

Explanation

To solve the problem step by step, we will calculate the power dissipated in the circuit consisting of an inductor and a resistor connected in series to an AC source.

### Step 1: Identify the given values
- Inductor reactance, \( X_L = 1 \, \Omega \)
- Resistor resistance, \( R = 2 \, \Omega \)
- RMS voltage, \( V_{rms} = 6 \, V \)

### Step 2: Calculate the total impedance \( Z \)
The impedance \( Z \) in an L-R circuit is given by the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values:
\[ Z = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \, \Omega \]

### Step 3: Calculate the RMS current \( I_{rms} \)
Using the formula for current:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{6}{\sqrt{5}} \, A \]

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

### Step 5: Calculate the average power \( P_{average} \)
The average power dissipated in the circuit is given by:
\[ P_{average} = V_{rms} \cdot I_{rms} \cdot \cos \phi \]
Substituting the values we have calculated:
\[ P_{average} = 6 \cdot \left(\frac{6}{\sqrt{5}}\right) \cdot \left(\frac{2}{\sqrt{5}}\right) \]
Calculating this step by step:
1. Calculate \( I_{rms} \cdot \cos \phi \):
\[ I_{rms} \cdot \cos \phi = \left(\frac{6}{\sqrt{5}}\right) \cdot \left(\frac{2}{\sqrt{5}}\right) = \frac{12}{5} \]
2. Now substitute back into the power equation:
\[ P_{average} = 6 \cdot \frac{12}{5} = \frac{72}{5} = 14.4 \, W \]

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

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