Question

In the figure shown, after the switch $'S'$ is turned from position $'A'$ to position $'B'$, the energy dissipated in the circuit in terms of capacitance $'C'$ and total charge $'Q'$ is:

• A. $\frac{3}{8}\frac{Q^2}{C}$
• B. $\frac{4}{3}\frac{Q^2}{C}$
• C. $\frac{1}{8}\frac{Q^2}{C}$
• D. $\frac{5}{8}\frac{Q^2}{C}$

Answer 1

Answer: (A)

$V_{i} = \frac{1}{2}CE^{2} $
$ V_{f} = \frac{\left(CE\right)^{2}}{2\times4c} = \frac{1}{2} \frac{CE^{2}}{4} $
$ \Delta E =\frac{1}{2} CE^{2} \times\frac{3}{4} = \frac{3}{8} CE^{2} $