The reactance of a capacitor of capacitance C is X. If both the frequency and capacitance be doubled, then new reactance will be
(a) `X` (b) `2X` (c) `4X` (d) `(X)/(4)`

Options

A. `X`
B. `2X`
C. `4X`
D. `(X)/(4)`

Answer

To solve the problem, we need to understand how the reactance of a capacitor changes when both the frequency and capacitance are altered. The reactance \( X \) of a capacitor is given by the formula:

\[ X = \frac{1}{\omega C} \]

where \( \omega \) is the angular frequency, and \( C \) is the capacitance.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- The initial capacitance is \( C \).
- The initial reactance is \( X \).
- The initial angular frequency \( \omega \) is related to the frequency \( f \) by \( \omega = 2\pi f \).

2. **Determine the new conditions**:
- Both the frequency and capacitance are doubled.
- The new capacitance \( C' = 2C \).
- The new frequency \( f' = 2f \), which means the new angular frequency \( \omega' = 2\omega \) (since \( \omega = 2\pi f \)).

3. **Write the expression for new reactance**:
- The new reactance \( X' \) can be expressed as:
\[ X' = \frac{1}{\omega' C'} \]

4. **Substitute the new values into the reactance formula**:
- Substitute \( \omega' = 2\omega \) and \( C' = 2C \):
\[ X' = \frac{1}{(2\omega)(2C)} = \frac{1}{4\omega C} \]

5. **Relate the new reactance to the old reactance**:
- Since the original reactance \( X = \frac{1}{\omega C} \), we can express \( X' \) in terms of \( X \):
\[ X' = \frac{1}{4} \cdot \frac{1}{\omega C} = \frac{X}{4} \]

6. **Conclusion**:
- Therefore, the new reactance \( X' \) is:
\[ X' = \frac{X}{4} \]

Thus, the answer is \( \frac{X}{4} \) (option d).