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

Answer

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 \( R \), with the current leading the voltage by \( 45^\circ \).

### Step-by-Step Solution:

1. **Understanding the Phase Angle**:
- Given that the current leads the voltage by \( 45^\circ \), we can denote this phase angle as \( \phi = 45^\circ \).

2. **Using the Tangent of the Phase Angle**:
- We know that the tangent of the phase angle in an RLC circuit is given by:
\[ \tan \phi = \frac{X_C - X_L}{R} \]
- Where \( X_C \) is the capacitive reactance and \( X_L \) is the inductive reactance.

3. **Substituting the Values**:
- Since \( \phi = 45^\circ \), we have:
\[ \tan 45^\circ = 1 \]
- Therefore, we can write:
\[ 1 = \frac{X_C - X_L}{R} \]
- This simplifies to:
\[ X_C - X_L = R \]

4. **Expressing Reactances**:
- The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
- The inductive reactance \( X_L \) is given by:
\[ X_L = \omega L \]
- Where \( \omega = 2\pi f \).

5. **Substituting Reactances into the Equation**:
- Now substituting \( X_C \) and \( X_L \) into the equation:
\[ \frac{1}{\omega C} - \omega L = R \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{1}{\omega C} = R + \omega L \]
- Taking the reciprocal:
\[ \omega C = \frac{1}{R + \omega L} \]

7. **Finding Capacitance \( C \)**:
- Finally, we can express \( C \) as:
\[ C = \frac{1}{\omega (R + \omega L)} \]
- Substituting \( \omega = 2\pi f \):
\[ C = \frac{1}{2\pi f (R + 2\pi f L)} \]

### Final Expression for Capacitance:
\[ C = \frac{1}{2\pi f (R + 2\pi f L)} \]