Question

In the circuit shown, switch is placed in position 1, till the capacitor is charged to half of the maximum possible charge in this situation. Now the switch $S$ is placed in position 2. The maximum energy lost by the circuit after switch S is placed in position 2 is

• A. $\frac{1}{2}{CE}^2$
• B. $\frac{1}{8}{CE}^2$
• C. $\frac{7}{8}{CE}^2$
• D. $\frac{9}{8}{CE}^2$

Answer 1

Answer: (A)

Initially when the capacitor is connected tto the upper circuit, it will gain energy $=\frac{1}{2}C{E}^2$

Later when the capacitor is connected tto the lower circuit, it will get connected to reverse polarity of the battery and lose all the energy.

Hence energy lost $=\frac{1}{2}C{E}^2$

Please note that after losing all the energy it will gain equal energy but with opposite polarity.