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)/(2 pi f (2 pi f L - R))`
B. `(1)/(2 pi f (2 pi f L + R))`
C. `(1)/( pi f (2 pi f L - R))`
D. `(1)/( pi f (2 pi f L + R))`

Explanation

To solve the problem, we need to find the value of capacitance \( C \) in a series LCR circuit where the current leads the voltage by \( 45^\circ \). Here are the steps to derive the formula for \( C \):

### Step 1: Understand the Phase Relationship
Given that the current leads the voltage by \( 45^\circ \), we can conclude that the circuit behaves in a capacitive manner. This implies that the capacitive reactance \( X_C \) is greater than the inductive reactance \( X_L \).

### Step 2: Use the Phase Angle
The phase angle \( \phi \) is given as \( 45^\circ \). We know that:
\[ \tan \phi = \frac{X_C - X_L}{R} \]
For \( \phi = 45^\circ \):
\[ \tan 45^\circ = 1 \]
Thus:
\[ X_C - X_L = R \]

### Step 3: Express Reactances
The reactances can be expressed as:
\[ X_C = \frac{1}{\omega C} \quad \text{and} \quad X_L = \omega L \]
Substituting these into the equation from Step 2 gives:
\[ \frac{1}{\omega C} - \omega L = R \]

### Step 4: Rearranging the Equation
Rearranging the equation to solve for \( C \):
\[ \frac{1}{\omega C} = R + \omega L \]
Taking the reciprocal:
\[ \omega C = \frac{1}{R + \omega L} \]
Thus:
\[ C = \frac{1}{\omega(R + \omega L)} \]

### Step 5: Substitute for \( \omega \)
We know that \( \omega = 2 \pi f \). Substituting this into the equation for \( C \):
\[ C = \frac{1}{2 \pi f (R + 2 \pi f L)} \]

### Final Expression
Thus, the value of capacitance \( C \) is given by:
\[ C = \frac{1}{2 \pi f (R + 2 \pi f L)} \]