A capacitor of capacitance `4muF` is charged to 80V and another capacitor of capacitance `6muF` is charged to 30V are connected to each other using zero resistance wires such that the positive plate of one capacitor is connected to the positive plate of the other. The energy lost by the `4muF` capacitor in the process in `X**10^(-4)J`. Find the value of X.

Explanation

To solve the problem, we need to find the energy lost by the 4μF capacitor when it is connected to the 6μF capacitor. Here’s a step-by-step breakdown of the solution:

### 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, \( V_1 = 80 V \)
- Capacitance of the second capacitor, \( C_2 = 6 \mu F = 6 \times 10^{-6} F \)
- Voltage of the second capacitor, \( V_2 = 30 V \)

### Step 2: Calculate the initial energy stored in the 4μF capacitor
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
For the 4μF capacitor:
\[ E_{initial} = \frac{1}{2} \times C_1 \times V_1^2 = \frac{1}{2} \times (4 \times 10^{-6}) \times (80)^2 \]
Calculating this:
\[ E_{initial} = \frac{1}{2} \times 4 \times 10^{-6} \times 6400 = 12.8 \times 10^{-3} J = 0.0128 J \]

### Step 3: Calculate the common voltage when both capacitors are connected
When the capacitors are connected in parallel (positive to positive and negative to negative), the common voltage \( V_{common} \) can be calculated using the formula:
\[ V_{common} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \]
Substituting the values:
\[ V_{common} = \frac{(4 \times 10^{-6} \times 80) + (6 \times 10^{-6} \times 30)}{4 \times 10^{-6} + 6 \times 10^{-6}} \]
Calculating the numerator:
\[ = \frac{(320 \times 10^{-6}) + (180 \times 10^{-6})}{10 \times 10^{-6}} = \frac{500 \times 10^{-6}}{10 \times 10^{-6}} = 50 V \]

### Step 4: Calculate the final energy stored in the 4μF capacitor
Now we calculate the energy stored in the 4μF capacitor after the connection:
\[ E_{final} = \frac{1}{2} C_1 V_{common}^2 = \frac{1}{2} \times (4 \times 10^{-6}) \times (50)^2 \]
Calculating this:
\[ E_{final} = \frac{1}{2} \times 4 \times 10^{-6} \times 2500 = 5 \times 10^{-3} J = 0.005 J \]

### Step 5: Calculate the energy lost by the 4μF capacitor
The energy lost by the 4μF capacitor is given by:
\[ \Delta E = E_{initial} - E_{final} \]
Substituting the values:
\[ \Delta E = 0.0128 - 0.005 = 0.0078 J \]

### Step 6: Convert the energy lost into the required format
The problem states that the energy lost is in the form \( X \times 10^{-4} J \):
\[ 0.0078 J = 78 \times 10^{-4} J \]
Thus, the value of \( X \) is \( 78 \).

### Final Answer
The value of \( X \) is \( 78 \).

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