A `100muF` capacitor is charged to `100 V`. After the charging, battery is disconnected. The capacitor is then connected in parallel to another capacitor. The final voltage is `20 V`. Calculate the capacity of second capacitor.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the initial charge on the first capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
Where:
- \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \)
- \( V = 100 \, V \)

Substituting the values:
\[ Q = 100 \times 10^{-6} \, F \times 100 \, V = 10^{-4} \, C = 10,000 \, \mu C \]

### Step 2: Understand the scenario after connecting the second capacitor
After the battery is disconnected, the charged capacitor is connected in parallel to another capacitor. The final voltage across both capacitors is given as \( V_{AB} = 20 \, V \).

### Step 3: Calculate the charge on the first capacitor after connecting in parallel
The charge on the first capacitor when the voltage is \( 20 \, V \) can be calculated as:
\[ Q_1' = C_1 \times V_{AB} \]
Where:
- \( C_1 = 100 \, \mu F = 100 \times 10^{-6} \, F \)
- \( V_{AB} = 20 \, V \)

Substituting the values:
\[ Q_1' = 100 \times 10^{-6} \, F \times 20 \, V = 2000 \, \mu C \]

### Step 4: Calculate the charge on the second capacitor
Let the capacitance of the second capacitor be \( C_2 \). The charge on the second capacitor when the voltage is \( 20 \, V \) is given by:
\[ Q_2' = C_2 \times V_{AB} = C_2 \times 20 \]

### Step 5: Apply charge conservation
The total charge before and after connecting the second capacitor must be conserved. Therefore:
\[ Q = Q_1' + Q_2' \]
Substituting the values we have:
\[ 10,000 \, \mu C = 2000 \, \mu C + C_2 \times 20 \]

### Step 6: Solve for \( C_2 \)
Rearranging the equation:
\[ C_2 \times 20 = 10,000 - 2000 \]
\[ C_2 \times 20 = 8000 \]
\[ C_2 = \frac{8000}{20} = 400 \, \mu F \]

### Final Answer
The capacitance of the second capacitor \( C_2 \) is \( 400 \, \mu F \).

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