A coil of inductive reactance `31 Omega` has a resistance of `8 ohm`. 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. 0.56
B. 0.64
C. 0.8
D. 0.33

Answer

To find the power factor of the given circuit, we will 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 a series circuit with inductive and capacitive reactance is given by:
\[ X = X_C - X_L \]
Substituting the values:
\[ X = 25 \, \Omega - 31 \, \Omega = -6 \, \Omega \]
(The negative sign indicates that the circuit is capacitive.)

### Step 3: Calculate the impedance \( Z \)
The impedance \( Z \) of the circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the known values:
\[ Z = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \Omega \]

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

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

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