In a series R, L, C circuit `X_(L)=10Omega, X_(c) = 4omega`and `R = 6Omega`. Find the power factor of the circuit

Options

A. `(1)/sqrt2)`
B. `(sqrt3)/(2)`
C. `1//2`
D. nonc of theco

Answer

To find the power factor of the given series RLC circuit, we will follow these steps:

### Step 1: Identify the given values
- Inductive reactance, \(X_L = 10 \, \Omega\)
- Capacitive reactance, \(X_C = 4 \, \Omega\)
- Resistance, \(R = 6 \, \Omega\)

### Step 2: Calculate the net reactance
The net reactance \(X\) in the circuit is given by:
\[ X = X_L - X_C \]
Substituting the values:
\[ X = 10 \, \Omega - 4 \, \Omega = 6 \, \Omega \]

### 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:
\[ Z = \sqrt{(6 \, \Omega)^2 + (6 \, \Omega)^2} \]
\[ Z = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \, \Omega \]

### Step 4: Calculate the power factor
The power factor (PF) is given by the formula:
\[ \text{Power Factor} = \frac{R}{Z} \]
Substituting the values we have:
\[ \text{Power Factor} = \frac{6 \, \Omega}{6\sqrt{2} \, \Omega} = \frac{1}{\sqrt{2}} \]

### Final Answer
The power factor of the circuit is:
\[ \text{Power Factor} = \frac{1}{\sqrt{2}} \]

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