A capacitor of capacitance value $1 \mu \mathrm{~F}$ is charges to 30 V and the battery is then disc

Question

A capacitor of capacitance value $1 \mu \mathrm{~F}$ is charges to 30 V and the battery is then disconnected. If it is connected across a $2 \mu \mathrm{~F}$ capacitor, the energy lost by the system is\

Professor PC Thomas Classes · Answer

Original charge\ $\mathrm{q}=\mathrm{CV}=\left(1 	imes 10^{-6} \mathrm{~F}ight) 	imes 30 \mathrm{~V}=30 	imes 10^{-6} \mathrm{C}$\ Original energy $=\frac{1}{2} \mathrm{~V}^{2} \mathrm{C}=\frac{1}{2} 	imes 900 	imes 10^{-6}$\ $=450 	imes 10^{-6} \mathrm{~J}=450 \mu \mathrm{~J}$\ After disconnecting from the battery it is connected to $2 \mu \mathrm{~F}$. The total capacitance is $\mathrm{C}_{1}+\mathrm{C}_{2}$\ ie., $1 	imes 10^{-6} \mathrm{~F}+2 	imes 10^{-6} \mathrm{~F}=3 	imes 10^{-6} \mathrm{~F}$\ $	herefore$ Final energy $=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} 	imes \frac{900 	imes 10^{-12}}{3 	imes 10^{-6}}$\ $	herefore$ Loss of the energy of the system $=450-150 \mu \mathrm{~J}$\