A body of capacity 4u Fis charged to 80V and another body of capacity 6 `mu` F is charged to 30V. When they are connected the energy lost by 4 `mu F `is

Options

A. A) `7.8 mJ`
B. B) `4.6 mJ`
C. C) `3.2 mJ`
D. D) `2.5 mJ`

Answer

To solve the problem step by step, we will calculate the initial energy stored in each capacitor, find the final voltage after they are connected, and then determine the energy lost by the 4 µF capacitor.

### Step 1: Calculate the initial energy of the 4 µF capacitor

The formula for the energy stored in a capacitor is given by:

\[ U = \frac{1}{2} C V^2 \]

For the 4 µF capacitor charged to 80 V:

\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} \, \text{F} \times (80 \, \text{V})^2 \]

Calculating this:

\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} \times 6400 \]
\[ U_1 = 2 \times 10^{-6} \times 6400 = 12.8 \times 10^{-3} \, \text{J} = 12.8 \, \text{mJ} \]

### Step 2: Calculate the initial energy of the 6 µF capacitor

For the 6 µF capacitor charged to 30 V:

\[ U_2 = \frac{1}{2} \times 6 \times 10^{-6} \, \text{F} \times (30 \, \text{V})^2 \]

Calculating this:

\[ U_2 = \frac{1}{2} \times 6 \times 10^{-6} \times 900 \]
\[ U_2 = 3 \times 10^{-6} \times 900 = 2.7 \times 10^{-3} \, \text{J} = 2.7 \, \text{mJ} \]

### Step 3: Calculate the final voltage after connecting both capacitors

When the capacitors are connected, the charge will redistribute, and we can find the final voltage \( V_f \) using the formula:

\[ V_f = \frac{Q_1 + Q_2}{C_1 + C_2} \]

Where:
- \( Q_1 = C_1 V_1 = 4 \times 10^{-6} \times 80 \)
- \( Q_2 = C_2 V_2 = 6 \times 10^{-6} \times 30 \)

Calculating \( Q_1 \) and \( Q_2 \):

\[ Q_1 = 4 \times 10^{-6} \times 80 = 320 \times 10^{-6} \, \text{C} \]
\[ Q_2 = 6 \times 10^{-6} \times 30 = 180 \times 10^{-6} \, \text{C} \]

Now, substituting \( Q_1 \) and \( Q_2 \) into the final voltage equation:

\[ V_f = \frac{320 \times 10^{-6} + 180 \times 10^{-6}}{4 \times 10^{-6} + 6 \times 10^{-6}} = \frac{500 \times 10^{-6}}{10 \times 10^{-6}} = 50 \, \text{V} \]

### Step 4: Calculate the final energy of the 4 µF capacitor

Now, we can calculate the final energy \( U_f \) of the 4 µF capacitor at the final voltage:

\[ U_f = \frac{1}{2} \times 4 \times 10^{-6} \times (50)^2 \]

Calculating this:

\[ U_f = \frac{1}{2} \times 4 \times 10^{-6} \times 2500 \]
\[ U_f = 2 \times 10^{-6} \times 2500 = 5 \times 10^{-3} \, \text{J} = 5 \, \text{mJ} \]

### Step 5: Calculate the energy lost by the 4 µF capacitor

The energy lost by the 4 µF capacitor is given by:

\[ \text{Energy lost} = U_1 - U_f \]

Substituting the values:

\[ \text{Energy lost} = 12.8 \, \text{mJ} - 5 \, \text{mJ} = 7.8 \, \text{mJ} \]

### Final Answer

The energy lost by the 4 µF capacitor is **7.8 mJ**.

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