Question 170 of 180

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 $\dfrac{C}{2}$. The energy loss in the process after the charge is distributed between the two capacitors is :

A$\dfrac{1}{2}C{V}_{0}^{2}$B$\dfrac{1}{3}C{V}_{0}^{2}$C$\dfrac{1}{4}C{V}_{0}^{2}$D$\dfrac{1}{6}C{V}_{0}^{2}$

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Answer: (D) $\dfrac{1}{6}C{V}_{0}^{2}$

Heat Loss is given by,
$H=\dfrac{{C}_{1}{C}_{2}}{2\left({C}_{1}+{C}_{2}\right)}{\left({V}_{1}-{V}_{2}\right)}^{2}$
${V}_{1}$ is initial voltage of capacitor $C$$={V}_{0}$
${V}_{2}$ is the initial voltage of capacitor $\dfrac{C}{2}$$=0$
$=\dfrac{c\times \dfrac{c}{2}}{2\left(c+\dfrac{c}{2}\right)}{\left({V}_{0}-0\right)}^{2}=\dfrac{C}{6}{V}_{0}^{2}$
$H=\dfrac{1}{6}{C}_{0}{V}_{0}^{2}$