An inductor of reactance 2`Omega` and a resistor of 4`Omega` are connected in series to the terminals of a 12 V (rms ) AC source. The power dissipated in the circuit is

Options

A. 8 W
B. 12 W
C. 14.4 W
D. 28.8 W

Answer

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

### Step 1: Identify the given values
- Inductive reactance (X_L) = 2 Ω
- Resistance (R) = 4 Ω
- RMS Voltage (V_RMS) = 12 V

### Step 2: Calculate the impedance (Z) of the circuit
The impedance (Z) in an RL circuit is given by the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values:
\[ Z = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \, \Omega \]

### Step 3: Calculate the RMS current (I_RMS)
Using Ohm's law for AC circuits, the RMS current can be calculated as:
\[ I_{RMS} = \frac{V_{RMS}}{Z} \]
Substituting the values:
\[ I_{RMS} = \frac{12}{2\sqrt{5}} = \frac{12}{2\sqrt{5}} = \frac{6}{\sqrt{5}} \, A \]

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

### Step 5: Calculate the power dissipated (P)
The power dissipated in the circuit can be calculated using the formula:
\[ P = V_{RMS} \cdot I_{RMS} \cdot \cos \phi \]
Substituting the values:
\[ P = 12 \cdot \frac{6}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = 12 \cdot \frac{12}{5} = \frac{144}{5} = 28.8 \, W \]

### Final Answer
The power dissipated in the circuit is **28.8 W**.

---