Three capacitors of capacitance 2μF, 4μFand 8μF are connected in parallel and a P.D of 6V is maintained across each capacitor, the total energy stored is?
- 2.52 x 10-6J
- 6.90 x 10-6J
-
2.52 x 10-4J
- 6.90 x 10-4J
The correct answer is: C
Explanation
C = C1 + C2 + C3
= 2 + 4 + 8 = 14μF
V = 6V
Energy = 1/2CV2
1/2 x 14 x 10-6 x 62
= 2.52 x 10-4 J
There is an explanation video available .