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 steps:

### Step 1: Calculate the Effective Capacitance
When capacitors are connected in series, the effective capacitance \( C_{\text{effective}} \) can be calculated using the formula:
\[ \frac{1}{C_{\text{effective}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Given \( C_1 = 10 \, \mu F \) and \( C_2 = 20 \, \mu F \):
\[ \frac{1}{C_{\text{effective}}} = \frac{1}{10} + \frac{1}{20} = \frac{2 + 1}{20} = \frac{3}{20} \]
Thus,
\[ C_{\text{effective}} = \frac{20}{3} \, \mu F \]

### Step 2: Calculate the Charge on the Capacitors
Using the formula \( Q = C \times V \), where \( V = 200 \, V \):
\[ Q = C_{\text{effective}} \times V = \frac{20}{3} \times 10^{-6} \times 200 \]
Calculating this gives:
\[ Q = \frac{20 \times 200}{3} \times 10^{-6} = \frac{4000}{3} \times 10^{-6} = \frac{4}{3} \times 10^{-3} \, C \]

### Step 3: Reconnect the Capacitors
After disconnecting from the voltage source, the capacitors are reconnected with their positive plates together and negative plates together. The charge on each capacitor will redistribute.

### Step 4: Define Charges on Capacitors
Let \( Q_0 = \frac{4}{3} \times 10^{-3} \, C \) be the initial charge on each capacitor. When they are reconnected:
- The charge on the 10 µF capacitor becomes \( Q_1 = Q_0 - Q \)
- The charge on the 20 µF capacitor becomes \( Q_2 = Q_0 + Q \)

### Step 5: Set Up the Voltage Equations
Since the capacitors are now in parallel, the voltage across each capacitor must be equal:
\[ \frac{Q_1}{C_1} = \frac{Q_2}{C_2} \]
Substituting the values:
\[ \frac{Q_0 - Q}{10 \times 10^{-6}} = \frac{Q_0 + Q}{20 \times 10^{-6}} \]
Cross-multiplying gives:
\[ 20(Q_0 - Q) = 10(Q_0 + Q) \]
Expanding and simplifying:
\[ 20Q_0 - 20Q = 10Q_0 + 10Q \]
\[ 10Q_0 = 30Q \implies Q = \frac{1}{3}Q_0 \]

### Step 6: Calculate Final Charges
Now we can find the final charges:
- For the 10 µF capacitor:
\[ Q_1 = Q_0 - Q = Q_0 - \frac{1}{3}Q_0 = \frac{2}{3}Q_0 \]
- For the 20 µF capacitor:
\[ Q_2 = Q_0 + Q = Q_0 + \frac{1}{3}Q_0 = \frac{4}{3}Q_0 \]

### Step 7: Calculate the Potential Differences
Now we can find the potential difference across each capacitor:
- For the 10 µF capacitor:
\[ V_1 = \frac{Q_1}{C_1} = \frac{\frac{2}{3}Q_0}{10 \times 10^{-6}} = \frac{\frac{2}{3} \times \frac{4}{3} \times 10^{-3}}{10 \times 10^{-6}} = \frac{8 \times 10^{-3}}{30 \times 10^{-6}} = \frac{8}{30} \times 10^{3} = \frac{4}{15} \times 10^{3} \approx 267 \, V \]

- For the 20 µF capacitor:
\[ V_2 = \frac{Q_2}{C_2} = \frac{\frac{4}{3}Q_0}{20 \times 10^{-6}} = \frac{\frac{4}{3} \times \frac{4}{3} \times 10^{-3}}{20 \times 10^{-6}} = \frac{16 \times 10^{-3}}{60 \times 10^{-6}} = \frac{16}{60} \times 10^{3} = \frac{4}{15} \times 10^{3} \approx 267 \, V \]

### Summary of Results
The potential difference across the 10 µF capacitor is approximately \( 267 \, V \), and across the 20 µF capacitor is approximately \( 133 \, V \).