Question

A capacitor of capacitance \( C \) and potential \( V \) has energy \( E \). It is connected to another capacitor of capacitance \( 2C \) and potential \( 2V \). Then the loss of energy is \( \frac{x}{3} E \), where \( x \) is \( \_\_\_\_ \).

Answer 1

Correct Answer: 2

The problem asks for the value of \(x\), where the loss of energy upon connecting two capacitors is given as \( \frac{x}{3}E \). Here, \(E\) is the initial energy of the first capacitor.

Concept Used:

The solution is based on the principles of energy storage in capacitors and the conservation of charge.

1. Energy of a Capacitor: The electrical potential energy stored in a capacitor of capacitance \(C\) charged to a potential \(V\) is given by:

\[ E = \frac{1}{2}CV^2 \]

2. Conservation of Charge: When capacitors are connected, charge is redistributed, but the total charge of the isolated system is conserved. The total charge is the sum of the initial charges on the capacitors.

3. Common Potential: When capacitors are connected in parallel, they reach a common potential \(V_{common}\), which is calculated as:

\[ V_{common} = \frac{\text{Total Charge}}{\text{Total Capacitance}} = \frac{Q_1 + Q_2}{C_1 + C_2} \]

4. Energy Loss: The loss of energy is the difference between the total initial energy and the total final energy of the system. This loss is dissipated as heat in the connecting wires.

\[ \Delta E_{loss} = E_{initial} - E_{final} \]

Step-by-Step Solution:

Step 1: Calculate the initial energy of the system.

The system initially consists of two separate capacitors.

• For the first capacitor: Capacitance \(C_1 = C\), Potential \(V_1 = V\). Its energy is \(E_1 = \frac{1}{2}C_1V_1^2 = \frac{1}{2}CV^2\). The problem states that this is \(E\).
• For the second capacitor: Capacitance \(C_2 = 2C\), Potential \(V_2 = 2V\). Its energy is \(E_2 = \frac{1}{2}C_2V_2^2 = \frac{1}{2}(2C)(2V)^2 = \frac{1}{2}(2C)(4V^2) = 4CV^2\).

The total initial energy is the sum of the energies of the two capacitors:

\[ E_{initial} = E_1 + E_2 = \frac{1}{2}CV^2 + 4CV^2 = \frac{9}{2}CV^2 \]

Step 2: Calculate the common potential after connecting the capacitors.

First, find the initial charge on each capacitor:

• \(Q_1 = C_1V_1 = CV\)
• \(Q_2 = C_2V_2 = (2C)(2V) = 4CV\)

When connected in parallel (assuming positive plates are connected together), the total charge is conserved:

\[ Q_{total} = Q_1 + Q_2 = CV + 4CV = 5CV \]

The total capacitance of the parallel combination is:

\[ C_{total} = C_1 + C_2 = C + 2C = 3C \]

The common potential is:

\[ V_{common} = \frac{Q_{total}}{C_{total}} = \frac{5CV}{3C} = \frac{5}{3}V \]

Step 3: Calculate the final energy of the system.

The final energy is the energy stored in the equivalent capacitor at the common potential:

\[ E_{final} = \frac{1}{2}C_{total}V_{common}^2 = \frac{1}{2}(3C)\left(\frac{5}{3}V\right)^2 \] \[ E_{final} = \frac{1}{2}(3C)\left(\frac{25}{9}V^2\right) = \frac{25}{6}CV^2 \]

Step 4: Calculate the loss of energy.

\[ \Delta E_{loss} = E_{initial} - E_{final} = \frac{9}{2}CV^2 - \frac{25}{6}CV^2 \]

To subtract, we find a common denominator:

\[ \Delta E_{loss} = \frac{27}{6}CV^2 - \frac{25}{6}CV^2 = \frac{2}{6}CV^2 = \frac{1}{3}CV^2 \]

Final Computation & Result:

Step 5: Express the energy loss in terms of \(E\) and find the value of \(x\).

We know that the initial energy of the first capacitor is \(E = \frac{1}{2}CV^2\).

We can rewrite the energy loss in terms of \(E\):

\[ \Delta E_{loss} = \frac{1}{3}CV^2 = \frac{2}{3} \left( \frac{1}{2}CV^2 \right) = \frac{2}{3}E \]

The problem states that the loss of energy is \( \frac{x}{3}E \). Comparing this with our result:

\[ \frac{x}{3}E = \frac{2}{3}E \]

This implies that \(x = 2\).

The value of \(x\) is 2.