`4muF` capacitor is charged to 150 V and another capacitor of `6muF` is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced.

Options

A. 0.005J
B. 0.500J
C. 0.003J
D. 1.003J

Explanation

To solve the problem, we will follow these steps:

### 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 \( C_1 = 4 \mu F \) charged to \( V_1 = 150 V \):
\[ Q_1 = C_1 \times V_1 = 4 \times 10^{-6} F \times 150 V = 6 \times 10^{-4} C \]

For the second capacitor \( C_2 = 6 \mu F \) charged to \( V_2 = 200 V \):
\[ Q_2 = C_2 \times V_2 = 6 \times 10^{-6} F \times 200 V = 1.2 \times 10^{-3} C \]

### Step 2: Calculate the total charge when connected
When the two capacitors are connected together, the total charge \( Q_{total} \) is:
\[ Q_{total} = Q_1 - Q_2 = 6 \times 10^{-4} C - 1.2 \times 10^{-3} C = -6 \times 10^{-4} C \]
(Note: The negative sign indicates that the second capacitor has more charge than the first.)

### Step 3: Calculate the equivalent capacitance
The equivalent capacitance \( C_{eq} \) when capacitors are connected in parallel is:
\[ C_{eq} = C_1 + C_2 = 4 \mu F + 6 \mu F = 10 \mu F = 10 \times 10^{-6} F \]

### Step 4: Calculate the final voltage across the capacitors
The final voltage \( V_f \) across the capacitors can be calculated using:
\[ V_f = \frac{Q_{total}}{C_{eq}} = \frac{-6 \times 10^{-4} C}{10 \times 10^{-6} F} = -60 V \]
Thus, the potential difference across them is \( 60 V \).

### Step 5: Calculate the initial energy stored in each capacitor
The energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
Calculating for the first capacitor:
\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} F \times (150 V)^2 = \frac{1}{2} \times 4 \times 10^{-6} \times 22500 = 0.045 J \]

Calculating for the second capacitor:
\[ U_2 = \frac{1}{2} \times 6 \times 10^{-6} F \times (200 V)^2 = \frac{1}{2} \times 6 \times 10^{-6} \times 40000 = 0.12 J \]

### Step 6: Calculate the total initial energy
\[ U_{initial} = U_1 + U_2 = 0.045 J + 0.12 J = 0.165 J \]

### Step 7: Calculate the final energy stored in the combined system
Using the final voltage:
\[ U_f = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} \times 10 \times 10^{-6} F \times (60 V)^2 = \frac{1}{2} \times 10 \times 10^{-6} \times 3600 = 0.018 J \]

### Step 8: Calculate the heat produced
The heat produced \( Q \) is the difference in initial and final energy:
\[ Q = U_{initial} - U_f = 0.165 J - 0.018 J = 0.147 J \]

### Final Results
- The potential difference across the capacitors is \( 60 V \).
- The heat produced is \( 0.147 J \).