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 objective is to determine the value of \(x\), given that the energy dissipated upon connecting two capacitors is \( \frac{x}{3}E \), where \(E\) represents the initial energy of the first capacitor.

Concept Used:

The solution relies on the principles of capacitor energy storage and charge conservation.

1. Capacitor Energy: The potential energy stored in a capacitor with capacitance \(C\) and voltage \(V\) is given by:

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

2. Charge Conservation: When capacitors are interconnected, charge is redistributed, but the total charge in an isolated system remains constant. The total charge is the sum of the initial charges.

3. Common Potential: In a parallel connection of capacitors, they attain a common potential \(V_{common}\), calculated as:

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

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

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

Step-by-Step Solution:

Step 1: Compute the initial energy of the system.

The system initially comprises two distinct capacitors.

• Capacitor 1: Capacitance \(C_1 = C\), Voltage \(V_1 = V\). Its energy is \(E_1 = \frac{1}{2}C_1V_1^2 = \frac{1}{2}CV^2\), which is defined as \(E\) in the problem.
• Capacitor 2: Capacitance \(C_2 = 2C\), Voltage \(V_2 = 2V\). Its energy is \(E_2 = \frac{1}{2}C_2V_2^2 = \frac{1}{2}(2C)(2V)^2 = 4CV^2\).

The total initial energy is the sum of individual energies:

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

Step 2: Calculate the common potential after capacitor connection.

Determine the initial charge on each capacitor:

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

Upon parallel connection (assuming positive terminals are joined), total charge is conserved:

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

The total capacitance in parallel 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: Compute the final energy of the system.

The final energy is 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 energy loss.

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

Using 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 energy loss in terms of \(E\) and determine \(x\).

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

Rewrite the energy loss using \(E\):

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

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

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

Therefore, \(x = 2\).

The value of \(x\) is 2.