A coil of inductive reactance `31 Omega` has a resistance of `8 Omega`. It is placed in series with a condenser of capacitive reactance `25 Omega`. The combination is connected to an `ac` source of `110 V`. The power factor of the circuit is

Options

A. 0.33
B. 0.56
C. 0.64
D. `0.80`

Answer

To find the power factor of the given circuit, we can follow these steps:

### Step 1: Identify the given values
- Inductive reactance, \( X_L = 31 \, \Omega \)
- Resistance, \( R = 8 \, \Omega \)
- Capacitive reactance, \( X_C = 25 \, \Omega \)

### Step 2: Calculate the net reactance
The net reactance \( X \) in the circuit can be calculated using the formula:
\[ X = X_L - X_C \]
Substituting the values:
\[ X = 31 \, \Omega - 25 \, \Omega = 6 \, \Omega \]

### Step 3: Calculate the impedance \( Z \)
The impedance \( Z \) of the circuit is given by:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values:
\[ Z = \sqrt{(8 \, \Omega)^2 + (6 \, \Omega)^2} \]
Calculating the squares:
\[ Z = \sqrt{64 + 36} = \sqrt{100} = 10 \, \Omega \]

### Step 4: Calculate the power factor \( \cos \phi \)
The power factor is given by the formula:
\[ \text{Power Factor} = \cos \phi = \frac{R}{Z} \]
Substituting the values:
\[ \text{Power Factor} = \frac{8 \, \Omega}{10 \, \Omega} = 0.8 \]

### Conclusion
The power factor of the circuit is \( 0.8 \).

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