A `20 mu F` capacitors is charged by a `30 V` `d.c` supply and then connected across an uncharged `50 mu F` capacitor. Calculate (i) the final potential diff. across the combinition (ii) initial and final energies.

Answer

`8.57 V ; 9xx10^(-3) J, 2.57xx10^(-3) J`

Explanation

Here `C_(1) = 20 mu F = 20xx10^(-6) F`
`V_(1) = 30 V, C_(2) = 50 mu F = 50xx10^(-6) F`.
`V_(2) = 0 , V = ?`
`V = (C_(1) V_(1) + C_(2) V_(2))/(C_(1) + C_(2)) = (20xx10^(-6)xx30)/((20+50) 10^(-6))`
`= (60)/(7) V = 8.57 V`
Initial energy, `U_(1) = (1)/(2) C_(1) V_(1)^(2)`
`= (1)/(2) (20xx10^(-6)) (30)^(2) = 90xx10^(-3)` joule
Finia energy, `U_(2) = (1)/(2) (C_(1) + C_(2)) V^(2)`
`= (1)/(2) (20+50) 10^(-6) ((60)/(7))^(2) = 2.57xx10^(-3)` joule
Loss of energy `= U_(1) - U_(2)`
`= 9xx10^(-3) - 2.57xx10^(-3) = 6.43xx10^(-3) J`
This loss of energy appears in the form of heat and e.m. radiations.