A `2muF` capacitor `C_1` is charge to a voltage 100 V and a `4muF capacitor `C_2` is charged to a voltasge `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 of energy loss when two capacitors are connected in parallel, we will follow these steps:

### Step 1: Calculate the initial charge on each capacitor.
For capacitor \( C_1 \):
- Capacitance \( C_1 = 2 \mu F = 2 \times 10^{-6} F \)
- Voltage \( V_1 = 100 V \)
- Charge \( Q_1 = C_1 \times V_1 = 2 \times 10^{-6} F \times 100 V = 200 \mu C = 200 \times 10^{-6} C \)

For capacitor \( C_2 \):
- Capacitance \( C_2 = 4 \mu F = 4 \times 10^{-6} F \)
- Voltage \( V_2 = 50 V \)
- Charge \( Q_2 = C_2 \times V_2 = 4 \times 10^{-6} F \times 50 V = 200 \mu C = 200 \times 10^{-6} C \)

### Step 2: Calculate the total charge when both capacitors are connected in parallel.
- Total charge \( Q_{total} = Q_1 + Q_2 = 200 \mu C + 200 \mu C = 400 \mu C = 400 \times 10^{-6} C \)

### Step 3: Calculate the equivalent capacitance of the parallel combination.
- The equivalent capacitance \( C_{eq} \) for capacitors in parallel is given by:
\[ C_{eq} = C_1 + C_2 = 2 \mu F + 4 \mu F = 6 \mu F = 6 \times 10^{-6} F \]

### Step 4: Calculate the final voltage across the parallel combination.
- The final voltage \( V_f \) across the capacitors can be calculated using the total charge and equivalent capacitance:
\[ V_f = \frac{Q_{total}}{C_{eq}} = \frac{400 \times 10^{-6} C}{6 \times 10^{-6} F} = \frac{400}{6} V \approx 66.67 V \]

### Step 5: Calculate the initial energy stored in each capacitor.
- The energy stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]

For \( C_1 \):
\[ U_1 = \frac{1}{2} \times 2 \times 10^{-6} F \times (100 V)^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 10000 = 10^{-2} J = 0.01 J \]

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

### Step 6: Calculate the total initial energy.
\[ U_{initial} = U_1 + U_2 = 0.01 J + 0.005 J = 0.015 J \]

### Step 7: Calculate the final energy stored in the equivalent capacitor.
\[ U_{final} = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} \times 6 \times 10^{-6} F \times (66.67 V)^2 \]
Calculating \( (66.67)^2 \):
\[ (66.67)^2 \approx 4444.44 \]
Thus,
\[ U_{final} = \frac{1}{2} \times 6 \times 10^{-6} \times 4444.44 \approx 1.33 \times 10^{-2} J = 0.0133 J \]

### Step 8: Calculate the loss of energy.
\[ \text{Loss of Energy} = U_{initial} - U_{final} = 0.015 J - 0.0133 J \approx 0.0017 J = 1.7 \times 10^{-3} J \]

### Final Answer:
The loss of energy due to the parallel connection is approximately \( 1.7 \times 10^{-3} J \).

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