Two capacitors of capacitances `3 muF` and `6 muF` are charged to a potential of `12 V` each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

Explanation

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

### Step 1: Calculate the charge on each capacitor before they are connected.

1. For the first capacitor (C1 = 3 µF, V1 = 12 V):
\[ Q_1 = C_1 \times V_1 = 3 \, \mu F \times 12 \, V = 36 \, \mu C \]

2. For the second capacitor (C2 = 6 µF, V2 = 12 V):
\[ Q_2 = C_2 \times V_2 = 6 \, \mu F \times 12 \, V = 72 \, \mu C \]

### Step 2: Determine the total charge when the capacitors are connected.

When the capacitors are connected with the positive plate of one connected to the negative plate of the other, the charges will rearrange. The total charge (Q_total) is given by:
\[ Q_{\text{total}} = Q_2 - Q_1 = 72 \, \mu C - 36 \, \mu C = 36 \, \mu C \]

### Step 3: Calculate the equivalent capacitance of the combination.

Since the capacitors are connected in series, the equivalent capacitance (C_eq) is given by:
\[ C_{\text{eq}} = C_1 + C_2 = 3 \, \mu F + 6 \, \mu F = 9 \, \mu F \]

### Step 4: Calculate the voltage across the equivalent capacitor.

Using the formula \( V = \frac{Q}{C} \):
\[ V = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{36 \, \mu C}{9 \, \mu F} = 4 \, V \]

### Conclusion

The potential difference across each capacitor after they are connected will be **4 V**.

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