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. `0.80`
D. `0.33`

Answer

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

### Step 1: Identify the given values
- Inductive reactance (X_L) = 31 Ω
- Resistance (R) = 8 Ω
- Capacitive reactance (X_C) = 25 Ω
- AC source voltage (V) = 110 V (not needed for power factor calculation)

### Step 2: Calculate the net reactance (X)
In a series RLC circuit, the net reactance (X) is given by:
\[ X = X_L - X_C \]
Substituting the values:
\[ X = 31 \, \Omega - 25 \, \Omega = 6 \, \Omega \]

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

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

### Conclusion
The power factor of the circuit is **0.8**.

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