The current in a circuit containing a capacitance `C` and a resistance `R` in series leads over the applied voltage of frequency `omega/(2pi)` by.

Options

A. `tan^(-1)(1)/(omegaCR)`
B. `tan^(-1)(omegaCR)`
C. `tan^(-1)(omega1)/R`
D. `cos^(-1) (omegaCR)`

Answer

To solve the problem of determining how much the current in a circuit containing a capacitance \( C \) and a resistance \( R \) in series leads over the applied voltage of frequency \( \frac{\omega}{2\pi} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
- The circuit consists of a resistor \( R \) and a capacitor \( C \) connected in series with an alternating current (AC) source.

2. **Identify the Voltage and Current Equations**:
- The voltage across the circuit can be represented as:
\[ V(t) = V_0 \sin(\omega t) \]
- The current in the circuit can be expressed as:
\[ I(t) = I_0 \sin(\omega t + \phi) \]
- Here, \( \phi \) is the phase angle by which the current leads or lags the voltage.

3. **Determine the Phase Relationship**:
- In an RC circuit, the current leads the voltage. The relationship between the voltages across the resistor \( V_R \) and the capacitor \( V_C \) can be analyzed using phasor diagrams.

4. **Use Phasor Diagram**:
- The phasor for the current \( I \) is taken as the reference phasor. The voltage across the resistor \( V_R \) is in phase with the current, while the voltage across the capacitor \( V_C \) lags the current by \( 90^\circ \).

5. **Apply the Relationships**:
- The relationship between the voltages can be expressed as:
\[ \tan(\theta) = \frac{V_C}{V_R} \]
- From the definitions, we know:
\[ V_R = I \cdot R \quad \text{and} \quad V_C = I \cdot X_C \]
- Here, \( X_C = \frac{1}{\omega C} \) is the capacitive reactance.

6. **Substituting Values**:
- Substitute \( V_R \) and \( V_C \) into the tangent equation:
\[ \tan(\theta) = \frac{I \cdot X_C}{I \cdot R} = \frac{X_C}{R} \]
- This simplifies to:
\[ \tan(\theta) = \frac{1}{\omega C R} \]

7. **Find the Phase Angle**:
- Therefore, the phase angle \( \theta \) can be expressed as:
\[ \theta = \tan^{-1}\left(\frac{1}{\omega C R}\right) \]

### Final Answer:
The current in the circuit leads over the applied voltage by:
\[ \theta = \tan^{-1}\left(\frac{1}{\omega C R}\right) \]