An inductor (L) and resistance (R) are connected in series with an AC source. The phase difference between voltage (V) and current (i) is `45^(@)`. Now a capacitor (C) is connected in series with L-R, If the phase difference between V and i remain same, then capacitive reactance and impedance of L-C-R circuit will be-

Options

A. `R//Rsqrt(2)`
B. `2R,Rsqrt(2)`
C. `R,R`
D. `2R,Rsqrt(3)`

Answer

To solve the problem, we need to determine the capacitive reactance (XC) and the impedance (Z) of the L-C-R circuit when the phase difference between voltage (V) and current (i) remains at 45 degrees after adding the capacitor.

### Step-by-Step Solution:

1. **Understanding the Phase Difference**:
The phase difference (φ) between the voltage and current in an L-R circuit is given as 45 degrees. This can be expressed mathematically as:
\[ \tan(φ) = \frac{X_L}{R} \]
where \(X_L\) is the inductive reactance and \(R\) is the resistance.

2. **Calculating Inductive Reactance**:
Since φ = 45 degrees, we have:
\[ \tan(45^\circ) = 1 \implies \frac{X_L}{R} = 1 \implies X_L = R \]

3. **Adding the Capacitor**:
When the capacitor (C) is connected in series with the L-R circuit, the new phase difference remains 45 degrees. The new relationship can be expressed as:
\[ \tan(φ) = \frac{X_C - X_L}{R} \]
where \(X_C\) is the capacitive reactance.

4. **Setting Up the Equation**:
Since the phase difference remains 45 degrees:
\[ \tan(45^\circ) = 1 \implies \frac{X_C - X_L}{R} = 1 \]
Substituting \(X_L = R\):
\[ \frac{X_C - R}{R} = 1 \]

5. **Solving for Capacitive Reactance**:
Rearranging the equation gives:
\[ X_C - R = R \implies X_C = 2R \]

6. **Calculating Impedance**:
The impedance (Z) of the L-C-R circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + (X_C - X_L)^2} \]
Substituting \(X_C = 2R\) and \(X_L = R\):
\[ Z = \sqrt{R^2 + (2R - R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]

### Final Answers:
- **Capacitive Reactance (XC)**: \(2R\)
- **Impedance (Z)**: \(R\sqrt{2}\)