A capacitorn of `2 muF` charged to 50 V is connected in parallel with another capacitor of `1 muF` charged to 20 V. The common potential and loss of energy will be

Options

A. `40 V,300 mu J`
B. `50 V,400 mu J`
C. `40 V, 600 mu J`
D. `50 V, 700 mu J`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial charges on both capacitors.
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the voltage.

For the first capacitor (2 µF charged to 50 V):
\[ Q_1 = C_1 \times V_1 = 2 \, \mu F \times 50 \, V = 100 \, \mu C \]

For the second capacitor (1 µF charged to 20 V):
\[ Q_2 = C_2 \times V_2 = 1 \, \mu F \times 20 \, V = 20 \, \mu C \]

### Step 2: Calculate the total charge when both capacitors are connected in parallel.
The total charge \( Q_{net} \) is the sum of the individual charges:
\[ Q_{net} = Q_1 + Q_2 = 100 \, \mu C + 20 \, \mu C = 120 \, \mu C \]

### Step 3: Calculate the equivalent capacitance of the two capacitors in parallel.
The equivalent capacitance \( C_{eq} \) for capacitors in parallel is given by:
\[ C_{eq} = C_1 + C_2 = 2 \, \mu F + 1 \, \mu F = 3 \, \mu F \]

### Step 4: Calculate the common potential across the capacitors.
The common potential \( V_{common} \) can be calculated using the formula:
\[ V_{common} = \frac{Q_{net}}{C_{eq}} = \frac{120 \, \mu C}{3 \, \mu F} = 40 \, V \]

### Step 5: Calculate the initial energy stored in both capacitors.
The energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
For the first capacitor:
\[ U_1 = \frac{1}{2} \times 2 \, \mu F \times (50 \, V)^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 2500 = 2.5 \, mJ = 2500 \, \mu J \]

For the second capacitor:
\[ U_2 = \frac{1}{2} \times 1 \, \mu F \times (20 \, V)^2 = \frac{1}{2} \times 1 \times 10^{-6} \times 400 = 0.2 \, mJ = 200 \, \mu J \]

Total initial energy:
\[ U_{initial} = U_1 + U_2 = 2500 \, \mu J + 200 \, \mu J = 2700 \, \mu J \]

### Step 6: Calculate the final energy stored in the equivalent capacitor.
Using the common potential:
\[ U_{final} = \frac{1}{2} C_{eq} V_{common}^2 = \frac{1}{2} \times 3 \, \mu F \times (40 \, V)^2 = \frac{1}{2} \times 3 \times 10^{-6} \times 1600 = 2.4 \, mJ = 2400 \, \mu J \]

### Step 7: Calculate the loss of energy.
The loss of energy \( \Delta U \) is given by:
\[ \Delta U = U_{initial} - U_{final} = 2700 \, \mu J - 2400 \, \mu J = 300 \, \mu J \]

### Final Answers:
- Common potential: **40 V**
- Loss of energy: **300 µJ**