Two capacitors of capacitances `3muF` and `6muF` are charged to a potential of `12V` each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other.
(A) the common potential will be `12V`
(B) the loss in energy will be `576muJ`
(c) the common potential will be `4V`
(D) the loss in energy will be `144 muJ`

Explanation

To solve the problem step by step, we need to analyze the situation with the two capacitors, their charges, and the energy before and after they are connected. Here’s how we can approach the problem:

### Step 1: Calculate the initial charge on each capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
For the first capacitor:
- Capacitance \( C_1 = 3 \mu F \)
- Voltage \( V_1 = 12 V \)
\[ Q_1 = C_1 \times V_1 = 3 \mu F \times 12 V = 36 \mu C \]

For the second capacitor:
- Capacitance \( C_2 = 6 \mu F \)
- Voltage \( V_2 = 12 V \)
\[ Q_2 = C_2 \times V_2 = 6 \mu F \times 12 V = 72 \mu C \]

### Step 2: Determine the net charge when connected
When the two capacitors are connected with opposite plates, the net charge \( Q_{net} \) will be:
\[ Q_{net} = Q_2 - Q_1 = 72 \mu C - 36 \mu C = 36 \mu C \]

### Step 3: Calculate the equivalent capacitance
Since the capacitors are connected in parallel, the equivalent capacitance \( C_{eq} \) is:
\[ C_{eq} = C_1 + C_2 = 3 \mu F + 6 \mu F = 9 \mu F \]

### Step 4: Calculate the common potential
The common potential \( V_f \) across the capacitors can be calculated using:
\[ V_f = \frac{Q_{net}}{C_{eq}} = \frac{36 \mu C}{9 \mu F} = 4 V \]

### Step 5: Calculate the initial potential energy
The initial potential energy \( U_{initial} \) stored in the capacitors can be calculated using the formula:
\[ U = \frac{1}{2} C V^2 \]
For both capacitors:
\[ U_{1} = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 3 \mu F \times (12 V)^2 = \frac{1}{2} \times 3 \times 10^{-6} \times 144 = 216 \mu J \]
\[ U_{2} = \frac{1}{2} C_2 V_2^2 = \frac{1}{2} \times 6 \mu F \times (12 V)^2 = \frac{1}{2} \times 6 \times 10^{-6} \times 144 = 432 \mu J \]
\[ U_{initial} = U_{1} + U_{2} = 216 \mu J + 432 \mu J = 648 \mu J \]

### Step 6: Calculate the final potential energy
The final potential energy \( U_{final} \) after they are connected can be calculated as:
\[ U_{final} = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} \times 9 \mu F \times (4 V)^2 = \frac{1}{2} \times 9 \times 10^{-6} \times 16 = 72 \mu J \]

### Step 7: Calculate the loss in energy
The loss in energy \( \Delta U \) is given by:
\[ \Delta U = U_{initial} - U_{final} = 648 \mu J - 72 \mu J = 576 \mu J \]

### Final Results
- The common potential after connection is \( 4 V \).
- The loss in energy is \( 576 \mu J \).

### Conclusion
The correct answers are:
- (A) the common potential will be \( 12V \) - **False**
- (B) the loss in energy will be \( 576 \mu J \) - **True**
- (C) the common potential will be \( 4V \) - **True**
- (D) the loss in energy will be \( 144 \mu J \) - **False**