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

Explanation

To solve the problem step by step, we will first find the common potential difference across the two capacitors when they are connected and then calculate the heat produced.

### Step 1: Identify the given values
- Capacitance of the first capacitor, \( C_1 = 4 \mu F = 4 \times 10^{-6} F \)
- Voltage of the first capacitor, \( V_1 = 150 V \)
- Capacitance of the second capacitor, \( C_2 = 6 \mu F = 6 \times 10^{-6} F \)
- Voltage of the second capacitor, \( V_2 = 200 V \)

### Step 2: Calculate the initial charges on each capacitor
The charge on a capacitor is given by the formula:
\[ Q = C \times V \]
- Charge on the first capacitor, \( Q_1 = C_1 \times V_1 = 4 \times 10^{-6} F \times 150 V = 6 \times 10^{-4} C \)
- Charge on the second capacitor, \( Q_2 = C_2 \times V_2 = 6 \times 10^{-6} F \times 200 V = 1.2 \times 10^{-3} C \)

### Step 3: Calculate the total charge when the capacitors are connected
When the two capacitors are connected, the total charge \( Q_{total} \) is the sum of the individual charges:
\[ Q_{total} = Q_1 - Q_2 = 6 \times 10^{-4} C - 1.2 \times 10^{-3} C = -6 \times 10^{-4} C \]
(Note: The negative sign indicates that the second capacitor has more charge, which will affect the final potential difference.)

### Step 4: Calculate the common potential difference
The common potential \( V_{common} \) across both capacitors can be calculated using the formula:
\[ V_{common} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \]
Substituting the values:
\[ V_{common} = \frac{(4 \times 10^{-6} F \times 150 V) + (6 \times 10^{-6} F \times 200 V)}{4 \times 10^{-6} F + 6 \times 10^{-6} F} \]
Calculating the numerator:
\[ = \frac{(6 \times 10^{-4} C) + (1.2 \times 10^{-3} C)}{10 \times 10^{-6} F} = \frac{1.8 \times 10^{-3} C}{10 \times 10^{-6} F} = 180 V \]

### Step 5: Calculate the initial potential energy
The initial potential energy \( U_{initial} \) is given by:
\[ U_{initial} = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 \]
Substituting the values:
\[ U_{initial} = \frac{1}{2} (4 \times 10^{-6} F)(150 V)^2 + \frac{1}{2} (6 \times 10^{-6} F)(200 V)^2 \]
Calculating each term:
\[ = \frac{1}{2} (4 \times 10^{-6} F)(22500 V^2) + \frac{1}{2} (6 \times 10^{-6} F)(40000 V^2) \]
\[ = 0.045 J + 0.12 J = 0.165 J \]

### Step 6: Calculate the final potential energy
The final potential energy \( U_{final} \) is given by:
\[ U_{final} = \frac{1}{2} (C_1 + C_2) V_{common}^2 \]
Substituting the values:
\[ U_{final} = \frac{1}{2} (10 \times 10^{-6} F)(180 V)^2 \]
Calculating:
\[ = \frac{1}{2} (10 \times 10^{-6} F)(32400 V^2) = 0.162 J \]

### Step 7: Calculate the heat produced
The heat produced \( Q_{heat} \) is the difference between the initial and final potential energy:
\[ Q_{heat} = U_{initial} - U_{final} = 0.165 J - 0.162 J = 0.003 J \]

### Final Answer
The potential difference across the capacitors is \( 180 V \) and the heat produced is \( 0.003 J \).