Question

a body of capacitor 6 micro farad is charged to 20V and another body of 4 micro farad is charged to 10V and both are connected with wire the energy lost by the system is
pleaase give answer with explanation​

Answer 1

Answer:

solution

Explanation:

Initial energy of body of capacitance 4μF is U

i

=1/2×4×10

−6

×80

2

=0.0128J

Final potential on this body after connection is

V=

4+6

4×80+6×30

=50V

so final energy on it,

U

f

=1/2×4×10

−6

×50

2

=0.005J Energy lost by this body=U

i

−U

f

=7.8mJ

Answer 2

Explanation:

1st one

= 6 micro for 20 v

so 6×20

=120

2nd one

=4 micro for 10V

So 4×10

=40

energy lost =

160+40

=200watt