A capacitor of `20muF` charged upto 500 V is connected in parallel with another capacitor of `10muF`, which is charged upto 200 V. the common potential is

Options

A. 200V
B. 300V
C. 400 V
D. 500 V

Explanation

To find the common potential when two capacitors are connected in parallel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitor 1: \( C_1 = 20 \, \mu F \) (microfarads), charged to \( V_1 = 500 \, V \) (volts)
- Capacitor 2: \( C_2 = 10 \, \mu F \), charged to \( V_2 = 200 \, V \)

2. **Understand the formula for common potential:**
- When capacitors are connected in parallel, the common potential \( V \) can be calculated using the formula:
\[ V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \]

3. **Substitute the values into the formula:**
- Plugging in the values:
\[ V = \frac{(20 \times 10^{-6} \, F) \times (500 \, V) + (10 \times 10^{-6} \, F) \times (200 \, V)}{(20 \times 10^{-6} \, F) + (10 \times 10^{-6} \, F)} \]

4. **Calculate the numerator:**
- Calculate \( C_1 V_1 \):
\[ C_1 V_1 = 20 \times 10^{-6} \times 500 = 10 \times 10^{-3} = 0.01 \, C \]
- Calculate \( C_2 V_2 \):
\[ C_2 V_2 = 10 \times 10^{-6} \times 200 = 2 \times 10^{-3} = 0.002 \, C \]
- Now, add these results:
\[ 0.01 + 0.002 = 0.012 \, C \]

5. **Calculate the denominator:**
- Calculate \( C_1 + C_2 \):
\[ C_1 + C_2 = 20 \times 10^{-6} + 10 \times 10^{-6} = 30 \times 10^{-6} \, F \]

6. **Calculate the common potential \( V \):**
- Now substitute the values back into the formula:
\[ V = \frac{0.012}{30 \times 10^{-6}} = \frac{0.012}{0.000030} = 400 \, V \]

### Final Answer:
The common potential when the two capacitors are connected in parallel is \( 400 \, V \).