Question

three capacitor 2uf,4ufand 8uf are joined in series and the combination is connected to a 12volt source.calculate the energy stored in the combination when these are fully charged . BRO PLEASE HELP ME ...MY EXAM IS MARCH-3 PLEASE HELP ME ​

Answer 1

Answer:

Mine exam is too on 3 march

Explanation:

Sum of Capacitor in series=1/c(sum)=1/c +1/c2 +1/c3

C(sum)= (4+2+1)/8

=7/8

Energy

=1*2(C)(V)2

=1/2*(7/8)* 144

=9*7

=63 Joule.

MARK BRAINLEST.