A `20muF` capacitor charged to 100 V is connected in parallel to a `10muF` capacitor charged to 100 V. find the loss in energy.

Answer

To find the loss in energy when a `20μF` capacitor charged to `100 V` is connected in parallel to a `10μF` capacitor also charged to `100 V`, we can follow these steps:

### 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 in farads and \( V \) is the voltage in volts.

**For the 20μF capacitor:**
\[ U_1 = \frac{1}{2} \times 20 \times 10^{-6} \, \text{F} \times (100 \, \text{V})^2 \]
\[ U_1 = \frac{1}{2} \times 20 \times 10^{-6} \times 10000 \]
\[ U_1 = \frac{1}{2} \times 20 \times 10^{-2} \]
\[ U_1 = 0.1 \, \text{J} \]

**For the 10μF capacitor:**
\[ U_2 = \frac{1}{2} \times 10 \times 10^{-6} \, \text{F} \times (100 \, \text{V})^2 \]
\[ U_2 = \frac{1}{2} \times 10 \times 10^{-6} \times 10000 \]
\[ U_2 = \frac{1}{2} \times 10 \times 10^{-2} \]
\[ U_2 = 0.05 \, \text{J} \]

### Step 2: Calculate the total initial energy.

The total initial energy \( U_{\text{initial}} \) is the sum of the energies stored in both capacitors:
\[ U_{\text{initial}} = U_1 + U_2 \]
\[ U_{\text{initial}} = 0.1 \, \text{J} + 0.05 \, \text{J} = 0.15 \, \text{J} \]

### Step 3: Determine the final voltage after connecting the capacitors in parallel.

When the capacitors are connected in parallel, the total charge \( Q \) is conserved, and the final voltage \( V_f \) can be calculated using the total capacitance:
\[ Q_1 = C_1 V_1 = 20 \times 10^{-6} \times 100 = 2 \times 10^{-3} \, \text{C} \]
\[ Q_2 = C_2 V_2 = 10 \times 10^{-6} \times 100 = 1 \times 10^{-3} \, \text{C} \]
\[ Q_{\text{total}} = Q_1 + Q_2 = 2 \times 10^{-3} + 1 \times 10^{-3} = 3 \times 10^{-3} \, \text{C} \]

The total capacitance \( C_{\text{total}} \) when connected in parallel is:
\[ C_{\text{total}} = C_1 + C_2 = 20 \times 10^{-6} + 10 \times 10^{-6} = 30 \times 10^{-6} \, \text{F} \]

Now, using \( Q_{\text{total}} = C_{\text{total}} V_f \):
\[ 3 \times 10^{-3} = 30 \times 10^{-6} V_f \]
\[ V_f = \frac{3 \times 10^{-3}}{30 \times 10^{-6}} = 100 \, \text{V} \]

### Step 4: Calculate the final energy stored in the combined capacitor.

The energy stored in the combined capacitor is:
\[ U_{\text{final}} = \frac{1}{2} C_{\text{total}} V_f^2 \]
\[ U_{\text{final}} = \frac{1}{2} \times 30 \times 10^{-6} \times (100)^2 \]
\[ U_{\text{final}} = \frac{1}{2} \times 30 \times 10^{-6} \times 10000 \]
\[ U_{\text{final}} = \frac{1}{2} \times 30 \times 10^{-2} \]
\[ U_{\text{final}} = 0.15 \, \text{J} \]

### Step 5: Calculate the loss in energy.

The loss in energy \( \Delta U \) is given by:
\[ \Delta U = U_{\text{initial}} - U_{\text{final}} \]
\[ \Delta U = 0.15 \, \text{J} - 0.15 \, \text{J} = 0 \, \text{J} \]

### Conclusion:

The loss in energy when the two capacitors are connected in parallel is \( 0 \, \text{J} \).

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