Question

Two identical capacitors have same capacitance C. One of them is charged to the potential V and other to the potential 2V. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is

• A. $\frac{CV^2}{4}$
• B. 2CV2
• C. $\frac{1}{2}CV^2$
• D. $\frac{3}{4}CV^2$

Answer 1

Answer: (A)

$\frac{CV^2}{4}$

Answer 2

When the capacitors are connected, the potential of the combined system is given by:

$V_c = \frac{q_{net}}{C_{net}} = \frac{CV + 2CV}{2C} = \frac{3V}{2}$

The loss of energy in the system can be calculated as:

$\text{Loss of energy} = \frac{1}{2}CV^2 + \frac{1}{2}C(2V)^2 - \frac{1}{2}C \left(\frac{3V}{2}\right)^2$

Simplifying this expression:

$\text{Loss of energy} = \frac{1}{2}CV^2 + \frac{1}{2}C(4V^2) - \frac{1}{2}C \left(\frac{9V^2}{4}\right)$

$\text{Loss of energy} = \frac{CV^2}{2} + 2CV^2 - \frac{9CV^2}{8}$

$\text{Loss of energy} = \frac{4CV^2}{8} + \frac{16CV^2}{8} - \frac{9CV^2}{8}$

$\text{Loss of energy} = \frac{CV^2}{4}$