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
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Correct Answer: Option 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
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