A `10 muC` capacitor and a `20 muF` capacitor are connected in series across a `200 V` supply line. The chraged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. what is the potential difference across each capacitor ?

Options

A. `(400)/(9) V`
B. `(800)/(9) V`
C. `400 V`
D. `200 V`

Explanation

To solve the problem step by step, let's break it down into manageable parts.

### Step 1: Understand the Initial Setup
We have two capacitors:
- Capacitor 1 (C1) = 10 µF
- Capacitor 2 (C2) = 20 µF

These capacitors are connected in series across a 200 V supply.

### Step 2: Calculate the Equivalent Capacitance
For capacitors in series, the equivalent capacitance (C_eq) can be calculated using the formula:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} \]
Thus,
\[ C_{eq} = \frac{20}{3} \, \mu F \approx 6.67 \, \mu F \]

### Step 3: Calculate the Charge on the Capacitors
The charge (Q) stored in the equivalent capacitor can be calculated using:
\[ Q = C_{eq} \times V \]
Substituting the values:
\[ Q = \left(\frac{20}{3} \times 10^{-6} \, F\right) \times 200 \, V = \frac{4000}{3} \, \mu C \approx 1333.33 \, \mu C \]

### Step 4: Charge on Each Capacitor
In a series connection, the charge on each capacitor is the same. Therefore, both capacitors have a charge of:
\[ Q_1 = Q_2 = Q = \frac{4000}{3} \, \mu C \]

### Step 5: Calculate the Voltage Across Each Capacitor
The voltage across each capacitor can be calculated using the formula:
\[ V = \frac{Q}{C} \]

For Capacitor 1 (C1):
\[ V_1 = \frac{Q}{C_1} = \frac{\frac{4000}{3} \, \mu C}{10 \, \mu F} = \frac{4000}{30} \, V \approx 133.33 \, V \]

For Capacitor 2 (C2):
\[ V_2 = \frac{Q}{C_2} = \frac{\frac{4000}{3} \, \mu C}{20 \, \mu F} = \frac{4000}{60} \, V \approx 66.67 \, V \]

### Step 6: Reconnect Capacitors in Parallel
After disconnecting from the supply, the capacitors are reconnected with their positive plates together and negative plates together. The total charge remains the same, and we can calculate the new voltage across each capacitor.

### Step 7: Calculate the New Common Voltage
The total charge when the capacitors are connected in parallel is:
\[ Q_{total} = Q_1 + Q_2 = \frac{4000}{3} + \frac{4000}{6} = \frac{8000}{6} \, \mu C = \frac{4000}{3} \, \mu C \]

The equivalent capacitance when connected in parallel:
\[ C_{parallel} = C_1 + C_2 = 10 \, \mu F + 20 \, \mu F = 30 \, \mu F \]

Using the total charge and equivalent capacitance, we can find the new common voltage (V):
\[ V = \frac{Q_{total}}{C_{parallel}} = \frac{\frac{4000}{3} \, \mu C}{30 \, \mu F} = \frac{4000}{90} \, V \approx 44.44 \, V \]

### Final Answer
The potential difference across each capacitor when they are reconnected in parallel is:
- Capacitor 1 (10 µF): 44.44 V
- Capacitor 2 (20 µF): 44.44 V