In a pure capacitive circuit if the frequency of ac source is doubled, then its capacitive reactance will be

Options

A. remains same
B. doubled
C. halved
D. zero

Explanation

To solve the problem of how the capacitive reactance changes when the frequency of an AC source is doubled in a pure capacitive circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Capacitive Reactance**:
The capacitive reactance \( X_c \) is given by the formula:
\[ X_c = \frac{1}{2 \pi f C} \]
where:
- \( X_c \) is the capacitive reactance,
- \( f \) is the frequency of the AC source,
- \( C \) is the capacitance.

2. **Initial Conditions**:
Let's denote the initial frequency as \( f \) and the initial capacitive reactance as \( X_c \).

3. **Doubling the Frequency**:
If the frequency is doubled, the new frequency \( f' \) becomes:
\[ f' = 2f \]

4. **Calculate New Capacitive Reactance**:
Substitute the new frequency \( f' \) into the formula for capacitive reactance:
\[ X_c' = \frac{1}{2 \pi f' C} = \frac{1}{2 \pi (2f) C} \]
Simplifying this gives:
\[ X_c' = \frac{1}{2 \cdot 2 \pi f C} = \frac{1}{2} \cdot \frac{1}{2 \pi f C} \]
Thus, we can express \( X_c' \) in terms of the original capacitive reactance \( X_c \):
\[ X_c' = \frac{X_c}{2} \]

5. **Conclusion**:
From the calculations, we find that when the frequency of the AC source is doubled, the new capacitive reactance \( X_c' \) is half of the original capacitive reactance \( X_c \):
\[ X_c' = \frac{X_c}{2} \]

### Final Answer:
The capacitive reactance will be halved.