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 calculate the initial energy stored in the capacitor, the final energy after connecting the two capacitors, and then determine the energy lost by the system.

### Step 1: Calculate the Initial Energy Stored in the Capacitor

The formula for the energy stored in a capacitor is given by:

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

Where:
- \( U \) is the energy,
- \( C \) is the capacitance,
- \( V \) is the voltage.

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

Substituting the values:

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

### Step 2: Determine the Charge on the Initial Capacitor

The charge \( Q \) on the capacitor can be calculated using the formula:

\[ Q = C \times V \]

Substituting the values:

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

### Step 3: Connect the Capacitors and Find the Final Voltage

When the charged capacitor is connected to a \( 2 \, \mu F \) capacitor, the total capacitance \( C_{total} \) is:

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

The total charge \( Q \) remains the same (conservation of charge):

\[ Q = 30 \, \mu C \]

The final voltage \( V_f \) across the capacitors can be found using:

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

### Step 4: Calculate the Final Energy Stored in the System

Now, we can calculate the final energy stored in the system using the new voltage and total capacitance:

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

Substituting the values:

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

### Step 5: Calculate the Energy Lost by the System

The energy lost \( \Delta U \) can be calculated as:

\[ \Delta U = U_{initial} - U_{final} \]

Substituting the values:

\[ \Delta U = 450 \, \mu J - 150 \, \mu J = 300 \, \mu J \]

### Final Answer

The energy lost by the system is \( 300 \, \mu J \).

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