In a circuit `L, C` and `R` are connected in series with an alternating voltage source of frequency `f`. The current lead the voltages by `45^(@)`. The value of `C` is :

Options

A. `1/(pif(2pifL+R))`
B. `1/(pif(2pifL-R))`
C. `1/(2pif(2pifL-R))`
D. `1/(2pif(2pifL+R))`

Explanation

To solve the problem, we need to find the value of capacitance \( C \) in a series circuit containing an inductor \( L \), a capacitor \( C \), and a resistor \( R \), given that the current leads the voltage by \( 45^\circ \).

### Step-by-Step Solution:

1. **Understanding the Phase Difference**:
The phase difference \( \phi \) between the current and voltage in an RLC circuit is given by the formula:
\[ \tan \phi = \frac{X_L - X_C}{R} \]
where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance.

2. **Substituting the Given Phase Angle**:
Since the current leads the voltage by \( 45^\circ \), we have:
\[ \tan 45^\circ = 1 \]
Therefore, we can write:
\[ 1 = \frac{X_L - X_C}{R} \]
This simplifies to:
\[ X_L - X_C = R \]

3. **Expressing Reactances**:
The reactances are defined as:
\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \) is the angular frequency.

4. **Substituting Reactances into the Equation**:
Substituting the expressions for \( X_L \) and \( X_C \) into the equation we derived:
\[ \omega L - \frac{1}{\omega C} = R \]
Replacing \( \omega \) with \( 2\pi f \):
\[ (2\pi f)L - \frac{1}{2\pi f C} = R \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ (2\pi f)L - R = \frac{1}{2\pi f C} \]

6. **Finding the Value of C**:
Taking the reciprocal of both sides:
\[ C = \frac{1}{2\pi f \left((2\pi f)L - R\right)} \]

7. **Final Expression**:
Thus, we can express \( C \) as:
\[ C = \frac{1}{2\pi f (2\pi f L - R)} \]

### Conclusion:
The value of capacitance \( C \) is given by:
\[ C = \frac{1}{2\pi f (2\pi f L - R)} \]