In series `LR` circuit, `X_(L) = 3 R`. Now a capacitor with `X_(C ) = R` is added in series. The ratio of new to old power factor

Options

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

Answer

To solve the problem step by step, we need to find the old and new power factors for the series LR circuit and then calculate the ratio of the new power factor to the old power factor.

### Step 1: Determine the Old Power Factor
In the original circuit, we have:
- Inductive reactance, \( X_L = 3R \)
- Resistance, \( R = R \)

The total impedance \( Z \) in the circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (3R)^2} = \sqrt{R^2 + 9R^2} = \sqrt{10R^2} = \sqrt{10}R \]

The power factor \( \phi_1 \) is defined as:
\[ \phi_1 = \frac{R}{Z} = \frac{R}{\sqrt{10}R} = \frac{1}{\sqrt{10}} \]

### Step 2: Determine the New Power Factor
Now, we add a capacitor with capacitive reactance \( X_C = R \) in series with the LR circuit. The new total impedance \( Z' \) becomes:
\[ Z' = \sqrt{R^2 + X_L^2 - X_C^2} = \sqrt{R^2 + (3R)^2 - R^2} = \sqrt{R^2 + 9R^2 - R^2} = \sqrt{9R^2} = 3R \]

Now, we can calculate the new power factor \( \phi_2 \):
\[ \phi_2 = \frac{R}{Z'} = \frac{R}{3R} = \frac{1}{3} \]

### Step 3: Calculate the Ratio of New to Old Power Factor
Now, we can find the ratio of the new power factor to the old power factor:
\[ \text{Ratio} = \frac{\phi_2}{\phi_1} = \frac{\frac{1}{3}}{\frac{1}{\sqrt{10}}} = \frac{1}{3} \cdot \frac{\sqrt{10}}{1} = \frac{\sqrt{10}}{3} \]

### Conclusion
Thus, the ratio of the new power factor to the old power factor is:
\[ \frac{\sqrt{10}}{3} \]