`A` and `B` are two condensers of capacities `2muF` and `4muF`. They are charged to potential differences of 12V and 6V respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is

Options

A. `24 muC` from A to B
B. `8muC` from A to B
C. `8mu C` from B to A
D. `24muC` from B to A

Explanation

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

### Step 1: Calculate the initial charges on each capacitor.
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the voltage.

1. For capacitor A:
- Capacitance \( C_1 = 2 \mu F = 2 \times 10^{-6} F \)
- Voltage \( V_1 = 12 V \)
- Charge \( Q_1 = C_1 \times V_1 = 2 \times 10^{-6} \times 12 = 24 \mu C \)

2. For capacitor B:
- Capacitance \( C_2 = 4 \mu F = 4 \times 10^{-6} F \)
- Voltage \( V_2 = 6 V \)
- Charge \( Q_2 = C_2 \times V_2 = 4 \times 10^{-6} \times 6 = 24 \mu C \)

### Step 2: Determine the total charge before connecting the capacitors.
The total charge \( Q_{total} \) before connecting the capacitors is:
\[ Q_{total} = Q_1 + Q_2 = 24 \mu C + 24 \mu C = 48 \mu C \]

### Step 3: Connect the capacitors (positive to positive).
When capacitors are connected positive to positive, they share the total charge until they reach a common potential. The total capacitance \( C_{net} \) of capacitors in parallel is given by:
\[ C_{net} = C_1 + C_2 = 2 \mu F + 4 \mu F = 6 \mu F \]

### Step 4: Calculate the common potential after connection.
The common potential \( V_{common} \) can be calculated using the formula:
\[ V_{common} = \frac{Q_{total}}{C_{net}} \]
Substituting the values:
\[ V_{common} = \frac{48 \mu C}{6 \mu F} = 8 V \]

### Step 5: Calculate the final charges on each capacitor.
Now that we have the common potential, we can find the final charges on each capacitor using the formula \( Q = C \times V \).

1. For capacitor A:
- \( Q_{A} = C_1 \times V_{common} = 2 \mu F \times 8 V = 16 \mu C \)

2. For capacitor B:
- \( Q_{B} = C_2 \times V_{common} = 4 \mu F \times 8 V = 32 \mu C \)

### Step 6: Calculate the charge that flows through the connecting wire.
The charge that flows through the connecting wire is the difference between the initial charge and the final charge on each capacitor.

1. For capacitor A:
- Initial charge \( Q_1 = 24 \mu C \)
- Final charge \( Q_{A} = 16 \mu C \)
- Charge that flows from A: \( Q_{flowA} = Q_1 - Q_{A} = 24 \mu C - 16 \mu C = 8 \mu C \)

2. For capacitor B:
- Initial charge \( Q_2 = 24 \mu C \)
- Final charge \( Q_{B} = 32 \mu C \)
- Charge that flows to B: \( Q_{flowB} = Q_{B} - Q_2 = 32 \mu C - 24 \mu C = 8 \mu C \)

### Final Answer:
The charge that flows through the connecting wire is \( 8 \mu C \).

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