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 step by step, we need to analyze the relationship between the AC source, the resistor, and the capacitor in the circuit.

### Step 1: Understand the Circuit
An AC source is connected in series with a resistor (R) and a capacitor (C). The voltage across each component will vary with time as the AC source provides a sinusoidal voltage.

### Step 2: Voltage Across the Resistor
The voltage across the resistor (V_R) in an AC circuit is given by:
\[ V_R = V_0 \sin(\omega t) \]
where \( V_0 \) is the maximum voltage and \( \omega \) is the angular frequency of the AC source.

### Step 3: Voltage Across the Capacitor
The voltage across the capacitor (V_C) lags behind the current by 90 degrees (or \( \frac{\pi}{2} \) radians). Therefore, the voltage across the capacitor can be expressed as:
\[ V_C = V_0 \sin(\omega t - \frac{\pi}{2}) \]
This can be rewritten using the sine function:
\[ V_C = V_0 \cos(\omega t) \]

### Step 4: Determine When the Voltages are in Phase
We need to find the time when the voltage across the resistor is in phase with the voltage across the capacitor. This means we need to find the time \( t \) when:
\[ V_R = V_C \]

From the equations derived:
\[ V_0 \sin(\omega t) = V_0 \cos(\omega t) \]

### Step 5: Simplify the Equation
Dividing both sides by \( V_0 \) (assuming \( V_0 \neq 0 \)):
\[ \sin(\omega t) = \cos(\omega t) \]

### Step 6: Use Trigonometric Identity
Using the identity \( \sin(x) = \cos(x) \) leads to:
\[ \tan(\omega t) = 1 \]

### Step 7: Solve for \( \omega t \)
The general solution for \( \tan(x) = 1 \) is:
\[ \omega t = \frac{\pi}{4} + n\pi \]
where \( n \) is an integer. The first positive solution occurs when \( n = 0 \):
\[ \omega t = \frac{\pi}{4} \]

### Step 8: Find Time \( t \)
To find \( t \):
\[ t = \frac{\pi}{4\omega} \]

### Step 9: Determine the First Instance of Phase Alignment
However, we are looking for the first instance when the voltage across the resistor is in phase with the initial potential difference across the capacitor. The first time this occurs is actually at:
\[ \omega t = \frac{3\pi}{2} \]

Thus:
\[ t = \frac{3\pi}{2\omega} \]

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

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