Question

A capacitor $C$ is fully charged with voltage $V_0$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}.$ The energy loss in the process after the charge is distributed between the two capacitors is :

• A. $\frac{1}{6}C{V}_0^2$
• B. $\frac{1}{2}C{V}_0^2$
• C. $\frac{1}{3}C{V}_0^2$
• D. $\frac{1}{4}C{V}_0^2$

Answer 1

Answer: (A)

$\frac{C{V}_0-q}{C}=\frac{q}{C/2}=\frac{2q}{C}$
$V_0=\frac{3q}{C}\Rightarrow q=\frac{C{V}_0}{3}$
$U_i=\frac{1}{2}C{V}_0^2$
$U_f=\frac{{\left(\frac{2C{V}_0}{3}\right)}^2}{2C}+\frac{{\left(\frac{C{V}_0}{3}\right)}^2}{2\left(\frac{C}{2}\right)}$
$=\frac{1}{2}C{V}_0^2\left[\frac{4}{9}+\frac{2}{9}\right]=\frac{1}{2}C{V}_0^2\left(\frac{2}{3}\right)$
Heat loss $=\frac{1}{2}C{V}_0^2-\left(\frac{2}{3}\right)\left(\frac{1}{2}C{V}_0^2\right)$
$=\frac{1}{6}C{V}_0^2$