For a series `RLC` circuit `R=X_(L)=2X_(C)`. The impedence of the current and phase different (between) `V` and `i` will be

Options

A. `(sqrt5R)/(2),tan^(-1)(2)`
B. `(sqrt5R)/(2),tan^(-1) ((1)/(2))`
C. `sqrt5X_(C),tan^(-1) ((1)/(2))`
D. `sqrt5R,tan^(-1) ((1)/(2))`

Answer

To solve the problem step by step, we will analyze the given conditions and apply the relevant formulas for impedance and phase difference in a series RLC circuit.

### Step 1: Understand the given conditions
We are given that:
- \( R = X_L = 2X_C \)

### Step 2: Express \( X_L \) and \( X_C \) in terms of \( R \)
From the given condition, we can express:
- \( X_L = R \)
- \( X_C = \frac{R}{2} \)

### Step 3: Write the formula for impedance \( Z \)
The impedance \( Z \) in a series RLC circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

### Step 4: Substitute the values of \( R \), \( X_L \), and \( X_C \)
Substituting \( X_L = R \) and \( X_C = \frac{R}{2} \) into the impedance formula:
\[ Z = \sqrt{R^2 + \left(R - \frac{R}{2}\right)^2} \]
This simplifies to:
\[ Z = \sqrt{R^2 + \left(\frac{R}{2}\right)^2} \]

### Step 5: Simplify the expression for \( Z \)
Calculating \( \left(\frac{R}{2}\right)^2 \):
\[ \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \]
Now substituting this back into the impedance formula:
\[ Z = \sqrt{R^2 + \frac{R^2}{4}} = \sqrt{\frac{4R^2}{4} + \frac{R^2}{4}} = \sqrt{\frac{5R^2}{4}} = \frac{R}{2} \sqrt{5} \]

### Step 6: Calculate the phase difference \( \phi \)
The phase difference \( \phi \) is given by:
\[ \tan \phi = \frac{X_L - X_C}{R} \]
Substituting the values:
\[ \tan \phi = \frac{R - \frac{R}{2}}{R} = \frac{\frac{R}{2}}{R} = \frac{1}{2} \]

### Step 7: Find \( \phi \)
To find \( \phi \), we take the arctangent:
\[ \phi = \tan^{-1}\left(\frac{1}{2}\right) \]

### Conclusion
The impedance of the circuit is \( \frac{R}{2} \sqrt{5} \) and the phase difference \( \phi \) is \( \tan^{-1}\left(\frac{1}{2}\right) \).