A series LR circuit is connected to an ac source of frequency `omega` and the inductive reactance is equal to 2R. A capacitance of capacitive reactance equal to R is added in series with L and R. The ratio of the new power factor to the old one is

Options

A. `sqrt((2)/(3))`
B. `sqrt((2)/(5))`
C. `sqrt((3)/(2))`
D. `sqrt((5)/(2))`

Explanation

To solve the problem step by step, we will analyze the series LR circuit and then the modified circuit with the added capacitor. We will calculate the power factors for both cases and find the ratio of the new power factor to the old one.

### Step 1: Analyze the original LR circuit
In the original LR circuit:
- The inductive reactance \( X_L \) is given as \( 2R \).
- The resistance \( R \) remains the same.

The impedance \( Z \) of the LR circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting \( X_L = 2R \):
\[ Z = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5} \]

### Step 2: Calculate the power factor for the original LR circuit
The power factor \( \text{pf} \) is defined as:
\[ \text{pf} = \cos \phi = \frac{R}{Z} \]
Substituting the values:
\[ \text{pf}_{\text{old}} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}} \]

### Step 3: Analyze the modified circuit with added capacitor
In the modified circuit:
- The capacitive reactance \( X_C \) is given as \( R \).
- The inductive reactance \( X_L \) remains \( 2R \).

The total reactance \( X \) in the modified circuit is:
\[ X = X_L - X_C = 2R - R = R \]

Now, we can calculate the new impedance \( Z' \):
\[ Z' = \sqrt{R^2 + X^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]

### Step 4: Calculate the power factor for the modified circuit
The new power factor \( \text{pf}' \) is:
\[ \text{pf}_{\text{new}} = \cos \phi' = \frac{R}{Z'} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Find the ratio of the new power factor to the old power factor
Now, we need to find the ratio of the new power factor to the old power factor:
\[ \text{Ratio} = \frac{\text{pf}_{\text{new}}}{\text{pf}_{\text{old}}} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{5}}} = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}} \]

### Final Answer
The ratio of the new power factor to the old power factor is:
\[ \sqrt{\frac{5}{2}} \]