Question

a capacitor of capacity 6microfarad is connected to a cell of 12V until it is fully charged .After disconnecting the cpacitor from then cell ,it is connected across another uncharged capacitor of cpacity 6microfarad .(i) calculate the energy loss during the combination of capacitors .

(ii) in which form this energy is dissipated?

Answer 1

$$\begin{gathered} \mathrm{Capacitance}\mathrm{of}\mathrm{a}\mathrm{charged}\mathrm{capacitor},{\mathrm{C}}_1=6\mathrm{\mu F}=6\times {10}^{-6}\mathrm{F} \\ \mathrm{Supply}\mathrm{voltage},{\mathrm{V}}_1=12\mathrm{V} \\ \mathrm{Electrostatic}\mathrm{energy}\mathrm{stored}\mathrm{in}{\mathrm{C}}_1, \\ {\mathrm{E}}_1=\frac{1}{2}{\mathrm{C}}_1{{\mathrm{V}}_1}^2 \\ =\frac{1}{2}\times 6\times {10}^{-6}\times 144=4.32\times {10}^{-4}\mathrm{J} \\ \mathrm{Capacitance}\mathrm{of}\mathrm{uncharged}\mathrm{capacitor},{\mathrm{C}}_2=6\mathrm{\mu F}=6\times {10}^{-6}\mathrm{F} \\ \mathrm{When}\mathrm{it}\mathrm{is}\mathrm{connected}\mathrm{to}\mathrm{the}\mathrm{circuit},\mathrm{the}\mathrm{potential}\mathrm{acquired}\mathrm{by}\mathrm{it}\mathrm{is}{\mathrm{V}}_2. \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{conservation}\mathrm{of}\mathrm{charge}, \\ {\mathrm{C}}_1{\mathrm{V}}_1={\mathrm{V}}_2({\mathrm{C}}_1+{\mathrm{C}}_2) \\ 6\times {10}^{-6}\times 12={\mathrm{V}}_2(6\times {10}^{-6}+6\times {10}^{-6})=6\times {10}^{-6}\times {\mathrm{V}}_2\times 2 \\ {\mathrm{V}}_2=\frac{6\times {10}^{-6}\times 12}{6\times {10}^{-6}\times 2}=6\mathrm{V} \\ \mathrm{Electrostatic}\mathrm{energy}\mathrm{for}\mathrm{the}\mathrm{combination}\mathrm{of}\mathrm{two}\mathrm{capacitors}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{E}}_2=\frac{1}{2}({\mathrm{C}}_1+{\mathrm{C}}_2){{\mathrm{V}}_2}^2 \\ =\frac{1}{2}\times (6\times {10}^{-6}+6\times {10}^{-6})\times 36 \\ =\frac{1}{2}\times 12\times {10}^{-6}\times 36=2.16\times {10}^{-4}\mathrm{J} \\ \mathrm{Therefore}, \\ \mathrm{Amount}\mathrm{of}\mathrm{electrostatic}\mathrm{energy}\mathrm{lost}\mathrm{by}\mathrm{capacitor} \\ {\mathrm{E}}_1-{\mathrm{E}}_2=4.32\times {10}^{-4}\mathrm{J}-2.16\times {10}^{-4}\mathrm{J}=2.16\times {10}^{-4}\mathrm{J} \end{gathered}$$
This energy is lost in the form of heat and electromagnetic radiation.