Question

A capacitor of capacity 4F is charged to SOV and
another capacitor of capacity 6F is charged to
30V. When they are connected. then the energy
lost by Apulf capacitori
(1) 7.8 m
(2) 46 m
() 3.2 ml
(4) 2.5 m​

Answer 1

Explanation:

Heat loss will be,

$heat = \frac{1}{2} \frac{c1 \times c2}{c1 + c2} {(v1 - v2)}^{2}$

Here, c1 =4F

c2= 6F

V1= 50V

V2= 30V

Solving, we will get heat loss=480 joule