To solve the problem, we will follow these steps:
### Step 1: Calculate the initial energy of the first capacitor (C1)
The formula for the energy stored in a capacitor is given by:
\[
U = \frac{1}{2} C V^2
\]
For the first capacitor:
- Capacitance \( C_1 = 4 \, \mu F = 4 \times 10^{-6} \, F \)
- Voltage \( V_1 = 50 \, V \)
Substituting the values:
\[
U_1 = \frac{1}{2} \times 4 \times 10^{-6} \times (50)^2
\]
Calculating:
\[
U_1 = \frac{1}{2} \times 4 \times 10^{-6} \times 2500 = 5 \times 10^{-3} \, J
\]
### Step 2: Calculate the initial energy of the second capacitor (C2)
For the second capacitor:
- Capacitance \( C_2 = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Voltage \( V_2 = 100 \, V \)
Substituting the values:
\[
U_2 = \frac{1}{2} \times 2 \times 10^{-6} \times (100)^2
\]
Calculating:
\[
U_2 = \frac{1}{2} \times 2 \times 10^{-6} \times 10000 = 1 \times 10^{-2} \, J
\]
### Step 3: Calculate the total initial energy (Ui)
Now, we add the energies of both capacitors:
\[
U_i = U_1 + U_2 = 5 \times 10^{-3} + 1 \times 10^{-2} = 1.5 \times 10^{-2} \, J
\]
### Step 4: Calculate the common voltage (Vc) after connection
When the capacitors are connected, the common voltage \( V_c \) can be calculated using the formula:
\[
V_c = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}
\]
Substituting the values:
\[
V_c = \frac{(4 \times 10^{-6} \times 50) + (2 \times 10^{-6} \times 100)}{4 \times 10^{-6} + 2 \times 10^{-6}}
\]
Calculating:
\[
V_c = \frac{(2 \times 10^{-4}) + (2 \times 10^{-4})}{6 \times 10^{-6}} = \frac{4 \times 10^{-4}}{6 \times 10^{-6}} = \frac{400}{6} \approx 66.67 \, V
\]
### Step 5: Calculate the final energy (Uf) after connection
The final energy stored in the combined capacitance is given by:
\[
U_f = \frac{1}{2} (C_1 + C_2) V_c^2
\]
Substituting the values:
\[
U_f = \frac{1}{2} \times (6 \times 10^{-6}) \times (66.67)^2
\]
Calculating:
\[
U_f = \frac{1}{2} \times 6 \times 10^{-6} \times 4444.44 \approx 1.33 \times 10^{-2} \, J
\]
### Step 6: Calculate the total energy before and after connection
Now we have:
- Initial energy \( U_i = 1.5 \times 10^{-2} \, J \)
- Final energy \( U_f = 1.33 \times 10^{-2} \, J \)
### Final Step: Express the energies in multiples of \( 10^{-2} \, J \)
- Initial energy in multiples of \( 10^{-2} \, J \): \( 1.5 \)
- Final energy in multiples of \( 10^{-2} \, J \): \( 1.33 \)
### Conclusion
The total energy before and after connection in multiples of \( 10^{-2} \, J \) is:
- Initial: \( 1.5 \)
- Final: \( 1.33 \)
