In a series LR circuit with `X_(L)=R`,power factor is `P_(1)`.If a capacitor of capacitance `C` with `X_(C)=X_(L)` is added to the circuit the power factor becomes `P_(2)`.The ratio of `P_(1)` to `P_(2)` will be:

Options

A. `1:2`
B. `1:3`
C. `1:1`
D. `1:sqrt(2)`

Answer

To solve the problem, we need to analyze the series LR circuit and the effect of adding a capacitor to it.

### Step-by-Step Solution:

1. **Understanding the Initial Circuit**:
- In the given series LR circuit, the inductive reactance \(X_L\) is equal to the resistance \(R\).
- Therefore, we have:
\[ X_L = R \]

2. **Calculating the Power Factor \(P_1\)**:
- The power factor \(P_1\) is defined as:
\[ P_1 = \cos(\theta_1) = \frac{R}{Z} \]
- Where \(Z\) is the impedance of the circuit. Since \(X_L = R\), we can express \(Z\) as:
\[ Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]
- Substituting \(Z\) into the power factor equation gives:
\[ P_1 = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} \]

3. **Adding the Capacitor**:
- When a capacitor with reactance \(X_C\) is added such that \(X_C = X_L\), we have:
\[ X_C = R \]
- The new impedance \(Z'\) of the circuit becomes:
\[ Z' = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (R - R)^2} = \sqrt{R^2} = R \]

4. **Calculating the New Power Factor \(P_2\)**:
- The new power factor \(P_2\) is given by:
\[ P_2 = \cos(\theta_2) = \frac{R}{Z'} = \frac{R}{R} = 1 \]

5. **Finding the Ratio of Power Factors**:
- Now, we can find the ratio of the two power factors:
\[ \frac{P_1}{P_2} = \frac{\frac{1}{\sqrt{2}}}{1} = \frac{1}{\sqrt{2}} \]

### Final Answer:
The ratio of \(P_1\) to \(P_2\) is:
\[ \frac{P_1}{P_2} = \frac{1}{\sqrt{2}} \]