An `8 (mu)F` capacitor is connected across a 220√2 V, 50 Hz line. What is the peak value of charge through the capacitor?

Options

A. (A) `2.5 xx10^(-3)C`
B. (B) `2.5 xx10^(-4)C`
C. (C) `5 xx10^(-5)C`
D. (D) `7.5 xx10^(-2)C`

Explanation

To solve the problem of finding the peak value of charge through an 8 µF capacitor connected across a 220√2 V, 50 Hz line, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance (C) = 8 µF = 8 × 10^(-6) F
- Voltage (V) = 220√2 V

2. **Calculate the peak voltage:**
- The peak voltage (V_peak) can be calculated as:
\[ V_{\text{peak}} = 220 \sqrt{2} \]
- Calculate \( 220 \sqrt{2} \):
\[ V_{\text{peak}} = 220 \times 1.414 \approx 311 \text{ V} \]

3. **Use the formula for charge in a capacitor:**
- The charge (Q) stored in a capacitor is given by the formula:
\[ Q = C \times V \]
- Substitute the values of C and V_peak into the formula:
\[ Q = (8 \times 10^{-6} \text{ F}) \times (311 \text{ V}) \]

4. **Calculate the charge:**
- Perform the multiplication:
\[ Q = 8 \times 311 \times 10^{-6} = 2488 \times 10^{-6} \text{ C} \]
- Simplifying gives:
\[ Q = 2.488 \times 10^{-3} \text{ C} \]

5. **Round the result:**
- Rounding to two significant figures, we get:
\[ Q \approx 2.5 \times 10^{-3} \text{ C} \]

### Final Answer:
The peak value of charge through the capacitor is approximately \( 2.5 \times 10^{-3} \) C.

---