An AC source is connected with a resistance ( R) and an unchanged capacitance C, in series. The potential difference across the resistor is in phase with the initial potential difference across the capacitor for the first time at the instant (assume that at t =0 , emf is zero)

Options

A. `pi/Omega`
B. `2pi/Omega`
C. `pi/2Omega`
D. `3pi/2Omega`

Explanation

To solve the problem, we need to analyze the relationship between the potential difference across the resistor (V_R) and the potential difference across the capacitor (V_C) in an AC circuit.

### Step-by-Step Solution:

1. **Understanding the AC Circuit**:
The AC source provides an alternating voltage, which can be expressed as:
\[ V(t) = V_0 \sin(\omega t) \]
where \( V_0 \) is the maximum voltage and \( \omega \) is the angular frequency.

2. **Voltage Across the Resistor**:
The voltage across the resistor (V_R) is in phase with the current. Therefore, we can express it as:
\[ V_R(t) = I(t) \cdot R \]
Since \( I(t) = \frac{V(t)}{Z} \) where \( Z \) is the impedance, and for a series RC circuit, the impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]
The current can be expressed as:
\[ I(t) = \frac{V_0}{Z} \sin(\omega t) \]
Thus, the voltage across the resistor becomes:
\[ V_R(t) = \frac{V_0}{Z} R \sin(\omega t) \]

3. **Voltage Across the Capacitor**:
The voltage across the capacitor (V_C) lags the current by \( \frac{\pi}{2} \) radians. Therefore, we can express it as:
\[ V_C(t) = \frac{1}{\omega C} I(t) = \frac{1}{\omega C} \cdot \frac{V_0}{Z} \sin(\omega t) \]
Since it lags by \( \frac{\pi}{2} \), we can write:
\[ V_C(t) = \frac{V_0}{Z \omega C} \sin\left(\omega t - \frac{\pi}{2}\right) = \frac{V_0}{Z \omega C} \cos(\omega t) \]

4. **Finding the Condition for In-Phase**:
We need to find the first time \( t \) when \( V_R(t) \) is in phase with \( V_C(t) \). This occurs when:
\[ V_R(t) = V_C(t) \]
Since \( V_R(t) \) is a sine function and \( V_C(t) \) is a cosine function, we can set:
\[ \sin(\omega t) = \cos(\omega t - \frac{\pi}{2}) \]
This means that:
\[ \sin(\omega t) = \sin(\omega t + \frac{\pi}{2}) \]
The sine function repeats every \( \pi \) radians, so we can find the first instance when they are equal.

5. **Solving for Time**:
The first time they are in phase after \( t = 0 \) is:
\[ \omega t = \frac{3\pi}{2} \]
Therefore, solving for \( t \):
\[ t = \frac{3\pi}{2\omega} \]

### Final Answer:
The potential difference across the resistor is in phase with the initial potential difference across the capacitor for the first time at:
\[ t = \frac{3\pi}{2\omega} \]