A conductor of capacitance `10mu F` connected to other conductor of capacitance `40mu F` having equal charges `100mu C` initially. Find out final voltage and heat loss during the process?

Explanation

`10 mu F " " 40 mu F`
`100mu C " " 100 mu C`
`C_(1)=10 mu F " " C_(2)=40 mu F`
`Q_(1)=100 mu C " " C_(2) = 100 mu C`
`V_(1)=Q_(1)//C_(1)=10V " " V_(2)=Q_(2)//C_(2)=2.5`
Final voltage (V) `=(C_(1)V_(1)+C(2)V_(2))/((C_(1)+C_(2))=(Q_(1)+Q_(2))/(C_(1)+C_(2))=(200mu C)/(50 mu F)=4V`
Heat loss during the process `=(1)/(2)[C_(1)C_(1)^(2)+C_(2)V_(2)^(2)]-(1)/(2)V^(2) (C_(1)+C_(2))`
`=(1)/(2)[Q_(1)V_(1)+Q_(2)V_(2)]-(1)/(2)V^(2)(C_(1)+C_(2))=(1)/(2)xx100mu [12.5]-(1)/(2)xx16(50) mu =225 mu J`