A capacitor has capacity C and reactance X. If capacitance and frequency become double, then reactance will be

Options

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

Answer

To solve the problem, we need to determine how the reactance of a capacitor changes when both its capacitance and the frequency of the AC source are doubled.

### Step-by-Step Solution:

1. **Understand the formula for capacitive reactance (X):**
The reactance \( X_C \) of a capacitor is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \) (angular frequency) and \( C \) is the capacitance.

2. **Identify the changes in capacitance and frequency:**
According to the problem, both the capacitance and frequency are doubled:
\[ C' = 2C \quad \text{and} \quad f' = 2f \]

3. **Calculate the new angular frequency:**
The new angular frequency \( \omega' \) is:
\[ \omega' = 2\pi f' = 2\pi (2f) = 4\pi f \]

4. **Substitute the new values into the reactance formula:**
Now, we can find the new reactance \( X_C' \):
\[ X_C' = \frac{1}{\omega' C'} = \frac{1}{(4\pi f)(2C)} \]

5. **Simplify the expression:**
\[ X_C' = \frac{1}{8\pi f C} \]

6. **Relate the new reactance to the original reactance:**
We know that the original reactance \( X_C \) is:
\[ X_C = \frac{1}{2\pi f C} \]
Therefore, we can express the new reactance in terms of the original reactance:
\[ X_C' = \frac{1}{8} \cdot \frac{1}{2\pi f C} = \frac{X_C}{4} \]

7. **Conclusion:**
Thus, the new reactance \( X_C' \) when both capacitance and frequency are doubled is:
\[ X_C' = \frac{X_C}{4} \]

### Final Answer:
The reactance will be \( \frac{X}{4} \).

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