A `2muF` capacitor `C_1` is charge to a voltage 100 V and a `4muF` capacitor `C_2` is charged to a voltage 50 V. The capacitors are then connected in parallel What is the loss of energy due to parallel connection?

Options

A. `1.7J`
B. `0.17J`
C. `1.7xx10^-2J`
D. `1.7xx10^-3J`

Explanation

To solve the problem, we need to determine the energy loss when two capacitors are connected in parallel. Here’s a step-by-step solution:

### Step 1: Calculate the initial energy stored in each capacitor

The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance and \( V \) is the voltage.

For capacitor \( C_1 \) (2 µF charged to 100 V):
\[ U_1 = \frac{1}{2} \times 2 \times 10^{-6} \, \text{F} \times (100 \, \text{V})^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 10000 = 10 \times 10^{-3} \, \text{J} = 0.01 \, \text{J} \]

For capacitor \( C_2 \) (4 µF charged to 50 V):
\[ U_2 = \frac{1}{2} \times 4 \times 10^{-6} \, \text{F} \times (50 \, \text{V})^2 = \frac{1}{2} \times 4 \times 10^{-6} \times 2500 = 5 \times 10^{-3} \, \text{J} = 0.005 \, \text{J} \]

### Step 2: Calculate the total initial energy

The total initial energy \( U_{\text{initial}} \) is the sum of the energies stored in both capacitors:
\[ U_{\text{initial}} = U_1 + U_2 = 0.01 \, \text{J} + 0.005 \, \text{J} = 0.015 \, \text{J} \]

### Step 3: Calculate the total charge and equivalent capacitance when connected in parallel

The total charge \( Q \) when the capacitors are connected in parallel is the sum of the individual charges:
\[ Q_1 = C_1 \times V_1 = 2 \times 10^{-6} \times 100 = 200 \times 10^{-6} \, \text{C} = 200 \, \mu\text{C} \]
\[ Q_2 = C_2 \times V_2 = 4 \times 10^{-6} \times 50 = 200 \times 10^{-6} \, \text{C} = 200 \, \mu\text{C} \]
\[ Q_{\text{total}} = Q_1 + Q_2 = 200 \, \mu\text{C} + 200 \, \mu\text{C} = 400 \, \mu\text{C} \]

The equivalent capacitance \( C_{\text{eq}} \) when connected in parallel is:
\[ C_{\text{eq}} = C_1 + C_2 = 2 \, \mu\text{F} + 4 \, \mu\text{F} = 6 \, \mu\text{F} \]

### Step 4: Calculate the final voltage across the capacitors

The final voltage \( V_f \) across the capacitors when connected in parallel is given by:
\[ V_f = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{400 \times 10^{-6}}{6 \times 10^{-6}} = \frac{400}{6} \approx 66.67 \, \text{V} \]

### Step 5: Calculate the final energy stored in the equivalent capacitor

The final energy \( U_{\text{final}} \) stored in the equivalent capacitor is:
\[ U_{\text{final}} = \frac{1}{2} C_{\text{eq}} V_f^2 = \frac{1}{2} \times 6 \times 10^{-6} \times (66.67)^2 \]
Calculating \( (66.67)^2 \):
\[ (66.67)^2 \approx 4444.44 \]
Thus,
\[ U_{\text{final}} = \frac{1}{2} \times 6 \times 10^{-6} \times 4444.44 \approx 13.33 \times 10^{-3} \, \text{J} = 0.01333 \, \text{J} \]

### Step 6: Calculate the energy loss

The energy loss \( \Delta U \) due to the parallel connection is:
\[ \Delta U = U_{\text{initial}} - U_{\text{final}} = 0.015 \, \text{J} - 0.01333 \, \text{J} \approx 0.00167 \, \text{J} \]

### Final Answer
The loss of energy due to the parallel connection is approximately:
\[ \Delta U \approx 1.67 \, \text{mJ} \]