A series `R-C` circuit is connected to an alternating voltage source. Consider two situations
(a) When capacitor is air filled.
(b) When capacitor is mica filled.
current through resistor is `i` and voltage across capacitor is `V` then

Options

A. `V_(a)=V_(b)`
B. `V_(a) lt V_(b)`
C. `V_(a )gt V_(b)`
D. `i_(a)=i_(b)`

Explanation

To solve the problem regarding the series R-C circuit connected to an alternating voltage source, we will analyze the two situations: when the capacitor is air-filled and when it is mica-filled.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have an R-C series circuit connected to an alternating voltage source (V).
- The current through the resistor is denoted as `i`, and the voltage across the capacitor is denoted as `V_C`.

2. **Capacitance of Capacitors**:
- For the air-filled capacitor (C_A), the capacitance is given by:
\[ C_A = \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.
- For the mica-filled capacitor (C_B), the capacitance is given by:
\[ C_B = \epsilon_r \cdot C_A = \epsilon_r \cdot \frac{\epsilon_0 A}{d} \]
where \( \epsilon_r > 1 \) (the relative permittivity of mica).

3. **Comparing Capacitances**:
- Since \( \epsilon_r > 1 \), it follows that:
\[ C_B > C_A \]

4. **Reactance of Capacitors**:
- The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
- Therefore, for the air-filled capacitor:
\[ X_{C_A} = \frac{1}{\omega C_A} \]
- And for the mica-filled capacitor:
\[ X_{C_B} = \frac{1}{\omega C_B} \]
- Since \( C_B > C_A \), it follows that \( X_{C_B} < X_{C_A} \).

5. **Impedance of the Circuit**:
- The total impedance \( Z \) of the circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
- For the air-filled capacitor:
\[ Z_A = \sqrt{R^2 + X_{C_A}^2} \]
- For the mica-filled capacitor:
\[ Z_B = \sqrt{R^2 + X_{C_B}^2} \]
- Since \( X_{C_B} < X_{C_A} \), it follows that \( Z_B < Z_A \).

6. **Current in the Circuit**:
- The current \( i \) in the circuit can be expressed as:
\[ i = \frac{V}{Z} \]
- Therefore:
\[ i_A = \frac{V}{Z_A} \quad \text{and} \quad i_B = \frac{V}{Z_B} \]
- Since \( Z_B < Z_A \), it follows that \( i_B > i_A \).

7. **Voltage Across the Capacitors**:
- The voltage across the capacitor is given by:
\[ V_C = i \cdot X_C \]
- For the air-filled capacitor:
\[ V_{C_A} = i_A \cdot X_{C_A} \]
- For the mica-filled capacitor:
\[ V_{C_B} = i_B \cdot X_{C_B} \]
- Since \( i_B > i_A \) and \( X_{C_B} < X_{C_A} \), we can conclude that:
\[ V_{C_B} < V_{C_A} \]

### Conclusion:
- Thus, we find that the voltage across the capacitor when it is air-filled is greater than when it is mica-filled:
\[ V_{C_A} > V_{C_B} \]

### Final Answer:
The correct option is that the voltage across the air-filled capacitor is greater than that across the mica-filled capacitor.

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