A capacitor of capacitance value `1 muF` is charged to 30 V and the battery is then disconnected. If it is connected across a `2 muF` capacotor, then the energy lost by the system is

Options

A. `300 mu J`
B. `450 muJ`
C. `225 mu J`
D. `150 muF`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the initial energy stored in the 1 µF capacitor
The energy \( U \) stored in a capacitor is given by the formula:

\[ U = \frac{1}{2} C V^2 \]

Where:
- \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)
- \( V = 30 \, V \)

Substituting the values:

\[ U = \frac{1}{2} \times (1 \times 10^{-6}) \times (30)^2 \]
\[ U = \frac{1}{2} \times (1 \times 10^{-6}) \times 900 \]
\[ U = \frac{900 \times 10^{-6}}{2} \]
\[ U = 450 \, \mu J \]

### Step 2: Calculate the charge on the 1 µF capacitor
The charge \( Q \) on the capacitor is given by:

\[ Q = C \times V \]

Substituting the values:

\[ Q = 1 \, \mu F \times 30 \, V \]
\[ Q = 1 \times 10^{-6} \, F \times 30 \, V = 30 \times 10^{-6} \, C = 30 \, \mu C \]

### Step 3: Connect the charged capacitor to the 2 µF capacitor
When the 1 µF capacitor is connected to the 2 µF capacitor, they will share the charge. The total capacitance \( C_{total} \) when connected in parallel is:

\[ C_{total} = C_1 + C_2 = 1 \, \mu F + 2 \, \mu F = 3 \, \mu F \]

### Step 4: Calculate the final voltage across both capacitors
Using the conservation of charge, the total charge before connection is equal to the total charge after connection:

\[ Q_{initial} = Q_{final} \]

Where:

\[ Q_{initial} = 30 \, \mu C \]

Let \( V_f \) be the final voltage across both capacitors:

\[ Q_{final} = C_{total} \times V_f \]
\[ 30 \, \mu C = 3 \, \mu F \times V_f \]

Solving for \( V_f \):

\[ V_f = \frac{30 \, \mu C}{3 \, \mu F} = 10 \, V \]

### Step 5: Calculate the final energy stored in the system
Now we calculate the energy stored in the system after the capacitors are connected:

\[ U_{final} = \frac{1}{2} C_{total} V_f^2 \]

Substituting the values:

\[ U_{final} = \frac{1}{2} \times 3 \, \mu F \times (10 \, V)^2 \]
\[ U_{final} = \frac{1}{2} \times 3 \times 10^{-6} \times 100 \]
\[ U_{final} = \frac{300 \times 10^{-6}}{2} \]
\[ U_{final} = 150 \, \mu J \]

### Step 6: Calculate the energy lost by the system
The energy lost \( \Delta U \) is the difference between the initial energy and the final energy:

\[ \Delta U = U_{initial} - U_{final} \]
\[ \Delta U = 450 \, \mu J - 150 \, \mu J \]
\[ \Delta U = 300 \, \mu J \]

### Final Answer:
The energy lost by the system is \( 300 \, \mu J \).

---