The reactance of a `25 muF` capacitor at the AC frequency of `4000Hz` is

Options

A. `5/(pi) ohm`
B. `sqrt(5/ (pi)) ohm`
C. `10 ohm`
D. `sqrt(10) ohm`

Explanation

To find the reactance of a 25 µF capacitor at an AC frequency of 4000 Hz, we can follow these steps:

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

### Step 2: Calculate the angular frequency
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2 \pi f \]
Given that \( f = 4000 \) Hz, we can calculate \( \omega \):
\[ \omega = 2 \pi \times 4000 \]

### Step 3: Substitute the values
Now, we can substitute \( \omega \) and \( C \) into the reactance formula. The capacitance \( C \) is given as 25 µF, which is:
\[ C = 25 \times 10^{-6} \text{ F} \]

### Step 4: Calculate the reactance
Substituting \( \omega \) and \( C \) into the reactance formula:
\[ X_C = \frac{1}{2 \pi \times 4000 \times 25 \times 10^{-6}} \]

### Step 5: Simplify the expression
Calculating the denominator:
\[ 2 \pi \times 4000 \times 25 \times 10^{-6} = 2 \pi \times 100 \times 10^{-3} = 200 \pi \times 10^{-3} \]
Thus,
\[ X_C = \frac{1}{200 \pi \times 10^{-3}} \]

### Step 6: Final calculation
Now, we can calculate \( X_C \):
\[ X_C = \frac{10^3}{200 \pi} = \frac{1000}{200 \pi} = \frac{5}{\pi} \text{ ohms} \]

### Final Answer
The reactance of the 25 µF capacitor at 4000 Hz is:
\[ X_C = \frac{5}{\pi} \text{ ohms} \approx 1.59 \text{ ohms} \]

---