3 capacitors of capacitance 15 µF, 30 µF and 35 µF are connected in parrellel to a 100v battery. calculate the total energy stored in the capacitors. calculate how much less energy would be stored if they were accidently connected in series.
Answer:
Energy = 1/2 * C * V^2
Total capacitance in parallel = 15 + 30 + 35 = 80 uF
Energy = 1/2 * 80e-6 * 100^2 = 0.4 Joules
If they are connected in series, the total capacitance is:
1 / (1/15 + 1/30 + 1/35) = 7.77 uF
1/2 * 7.77e-6 *100^2 = 0.039 Joules
.
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