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 follow these steps:

### Step 1: Calculate the initial energy stored in the 4μF capacitor.
The formula for the energy stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
For the 4μF capacitor charged to 80V:
\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} \, \text{F} \times (80 \, \text{V})^2 \]
Calculating this:
\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} \times 6400 = \frac{1}{2} \times 25600 \times 10^{-6} = 12800 \times 10^{-6} \, \text{J} = 0.0128 \, \text{J} \]

### Step 2: Calculate the initial energy stored in the 6μF capacitor.
For the 6μF capacitor charged to 30V:
\[ U_2 = \frac{1}{2} \times 6 \times 10^{-6} \, \text{F} \times (30 \, \text{V})^2 \]
Calculating this:
\[ U_2 = \frac{1}{2} \times 6 \times 10^{-6} \times 900 = \frac{1}{2} \times 5400 \times 10^{-6} = 2700 \times 10^{-6} \, \text{J} = 0.0027 \, \text{J} \]

### Step 3: Calculate the total charge on both capacitors.
The charge on the 4μF capacitor:
\[ Q_1 = C_1 V_1 = 4 \times 10^{-6} \, \text{F} \times 80 \, \text{V} = 320 \times 10^{-6} \, \text{C} \]
The charge on the 6μF capacitor:
\[ Q_2 = C_2 V_2 = 6 \times 10^{-6} \, \text{F} \times 30 \, \text{V} = 180 \times 10^{-6} \, \text{C} \]
Total charge when connected in parallel:
\[ Q_{total} = Q_1 + Q_2 = 320 \times 10^{-6} + 180 \times 10^{-6} = 500 \times 10^{-6} \, \text{C} \]

### Step 4: Calculate the equivalent capacitance.
When connected in parallel, the equivalent capacitance \(C_{eq}\) is:
\[ C_{eq} = C_1 + C_2 = 4 \times 10^{-6} + 6 \times 10^{-6} = 10 \times 10^{-6} \, \text{F} \]

### Step 5: Calculate the common potential across the capacitors.
Using the total charge and equivalent capacitance:
\[ V_{common} = \frac{Q_{total}}{C_{eq}} = \frac{500 \times 10^{-6}}{10 \times 10^{-6}} = 50 \, \text{V} \]

### Step 6: Calculate the final energy stored in the 4μF capacitor.
Using the common potential:
\[ U_1' = \frac{1}{2} C_1 V_{common}^2 = \frac{1}{2} \times 4 \times 10^{-6} \times (50)^2 \]
Calculating this:
\[ U_1' = \frac{1}{2} \times 4 \times 10^{-6} \times 2500 = \frac{1}{2} \times 10000 \times 10^{-6} = 5000 \times 10^{-6} \, \text{J} = 0.005 \, \text{J} \]

### Step 7: Calculate the energy lost by the 4μF capacitor.
The energy lost is:
\[ \Delta U = U_1 - U_1' = 0.0128 - 0.005 = 0.0078 \, \text{J} \]

### Step 8: Convert the energy lost into the required format.
To express this in the form \(X \times 10^{-4} \, \text{J}\):
\[ 0.0078 \, \text{J} = 78 \times 10^{-4} \, \text{J} \]
Thus, \(X = 78\).

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