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)ltV_(b)`
B. `V_(a)gtV_(b)`
C. `i_(a)gti_(b)`
D. `V_(a)=V_(b)`

Answer

To solve the problem regarding the series R-C circuit connected to an alternating voltage source under two different conditions (air-filled capacitor and mica-filled capacitor), we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a series R-C circuit connected to an AC voltage source. The circuit consists of a resistor (R) and a capacitor (C).

2. **Capacitive Reactance**:
- The capacitive reactance (X_C) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
- Where \( \omega \) is the angular frequency of the AC source and \( C \) is the capacitance of the capacitor.

3. **Impedance of the Circuit**:
- The total impedance (Z) of the R-C circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]

4. **Current through the Circuit**:
- The current (I) through the circuit can be expressed as:
\[ I = \frac{V}{Z} \]
- Where \( V \) is the voltage of the AC source.

5. **Case A: Air-Filled Capacitor**:
- Let’s denote the current through the air-filled capacitor as \( I_A \) and the voltage across the capacitor as \( V_A \).
- The capacitance \( C_A \) for an air-filled capacitor is simply \( C \).
- Thus, the capacitive reactance is:
\[ X_{C_A} = \frac{1}{\omega C} \]

6. **Case B: Mica-Filled Capacitor**:
- For the mica-filled capacitor, let’s denote the current as \( I_B \) and the voltage as \( V_B \).
- The capacitance \( C_B \) for a mica-filled capacitor is given by:
\[ C_B = K \cdot C \]
- Where \( K \) is the dielectric constant of mica (greater than 1).
- Therefore, the new capacitive reactance becomes:
\[ X_{C_B} = \frac{1}{\omega C_B} = \frac{1}{\omega K C} \]

7. **Comparing Impedances**:
- Since \( C_B > C_A \), it follows that \( X_{C_B} < X_{C_A} \).
- Consequently, the impedance \( Z_B < Z_A \).

8. **Effect on Current**:
- Since the impedance decreases in case B, the current increases:
\[ I_B > I_A \]

9. **Voltage Across the Capacitor**:
- The voltage across the capacitor can be expressed as:
\[ V = I \cdot X_C \]
- Therefore, for the air-filled capacitor:
\[ V_A = I_A \cdot X_{C_A} \]
- And for the mica-filled capacitor:
\[ V_B = I_B \cdot X_{C_B} \]

10. **Conclusion**:
- Since \( I_B > I_A \) and \( X_{C_B} < X_{C_A} \), it follows that:
\[ V_B < V_A \]
- Thus, the voltage across the mica-filled capacitor is less than that across the air-filled capacitor:
\[ V_B < V_A \]

### Final Answer:
The correct conclusion is that the voltage across the mica-filled capacitor (V_B) is less than the voltage across the air-filled capacitor (V_A).