Two coils `A` and `B` are connected in series across a `240V ,50 `Hz supply The resistance of A is `5Omega` and the inductance of `B` is 0.02 H The power factor is `0.75` The impedance of the circuit is (if power consumed is 3kW) .

Options

A. `0.144 Omega`
B. `1.44Omega`
C. `14.4 Omega`s
D. `144Omega`

Explanation

To find the impedance of the circuit with the given parameters, we can follow these steps:

### Step 1: Understand the given data
- Voltage (V) = 240 V
- Frequency (f) = 50 Hz
- Resistance of coil A (R) = 5 Ω
- Inductance of coil B (L) = 0.02 H
- Power factor (cos φ) = 0.75
- Power consumed (P) = 3 kW = 3000 W

### Step 2: Calculate the current (I) using the power formula
The power consumed in an AC circuit can be expressed as:
\[ P = V \cdot I \cdot \cos \phi \]
Rearranging this formula to find the current (I):
\[ I = \frac{P}{V \cdot \cos \phi} \]

Substituting the values:
\[ I = \frac{3000 \, \text{W}}{240 \, \text{V} \cdot 0.75} \]
\[ I = \frac{3000}{180} = 16.67 \, \text{A} \]

### Step 3: Calculate the inductive reactance (X_L)
The inductive reactance (X_L) can be calculated using the formula:
\[ X_L = 2 \pi f L \]
Substituting the values:
\[ X_L = 2 \pi \cdot 50 \, \text{Hz} \cdot 0.02 \, \text{H} \]
\[ X_L = 2 \cdot 3.14 \cdot 50 \cdot 0.02 \]
\[ X_L = 6.28 \]

### Step 4: Calculate the total impedance (Z)
The total impedance in a series circuit can be calculated using:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values:
\[ Z = \sqrt{(5 \, \Omega)^2 + (6.28 \, \Omega)^2} \]
\[ Z = \sqrt{25 + 39.4384} \]
\[ Z = \sqrt{64.4384} \]
\[ Z \approx 8.03 \, \Omega \]

### Step 5: Verify with the power factor
We can also verify the impedance using the power factor:
\[ Z = \frac{V}{I \cdot \cos \phi} \]
Substituting the values:
\[ Z = \frac{240 \, \text{V}}{16.67 \, \text{A} \cdot 0.75} \]
\[ Z = \frac{240}{12.5025} \]
\[ Z \approx 19.2 \, \Omega \]

### Conclusion
The impedance of the circuit is approximately \( Z \approx 19.2 \, \Omega \).