A `10 muF` 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

Answer

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

### Step 1: Calculate the Equivalent Capacitance
When capacitors are connected in series, the equivalent capacitance \(C_{eq}\) can be calculated using the formula:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
where \(C_1 = 10 \mu F\) and \(C_2 = 20 \mu F\).

Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{10} + \frac{1}{20} \]
\[ \frac{1}{C_{eq}} = \frac{2 + 1}{20} = \frac{3}{20} \]
Thus,
\[ C_{eq} = \frac{20}{3} \mu F \]

### Step 2: Calculate the Charge on the Capacitors
The charge \(Q\) stored in the capacitors when connected to a voltage \(V = 200 V\) is given by:
\[ Q = C_{eq} \times V \]
Substituting the values:
\[ Q = \left(\frac{20}{3} \mu F\right) \times 200 V = \frac{4000}{3} \mu C \]

### Step 3: Reconnect the Capacitors
After disconnecting from the voltage source, the capacitors are reconnected with their positive plates together and negative plates together. In this configuration, the total charge remains the same, and the capacitors are now in parallel.

### Step 4: Calculate the Total Charge
Since both capacitors have the same charge \(Q\) when they were in series, we can say:
\[ Q_1 = Q_2 = \frac{4000}{3} \mu C \]
Thus, the total charge \(Q_{total}\) when connected in parallel is:
\[ Q_{total} = Q_1 + Q_2 = \frac{4000}{3} + \frac{4000}{3} = \frac{8000}{3} \mu C \]

### Step 5: Calculate the New Equivalent Capacitance
When connected in parallel, the equivalent capacitance \(C_{eq}'\) is the sum of the individual capacitances:
\[ C_{eq}' = C_1 + C_2 = 10 \mu F + 20 \mu F = 30 \mu F \]

### Step 6: Calculate the Potential Difference Across Each Capacitor
The potential difference \(V\) across the capacitors can be calculated using:
\[ V = \frac{Q_{total}}{C_{eq}'} \]
Substituting the values:
\[ V = \frac{\frac{8000}{3} \mu C}{30 \mu F} = \frac{8000}{90} V = \frac{800}{9} V \]

### Final Result
Thus, the potential difference across each capacitor when they are reconnected is:
\[ V \approx 88.89 V \]