Question

Three capacitors 5uF, 3uF and 2uF are connected in series to a 100V d.c. Draw the circuit diagram.
Calculate the energy stored by the system

Answer 1

for each
C = q/V so V = q/C
They each have the same charge q
V1 = q / (5 *10^-6)
V2 = q / (3 *10^-6)
V3 = q / (2 *10^-6)
so
10 = 10^6 q [ 1 / 5 +1/3 + 1/2 ] =10^6 q * 31/30
q = 10^-6 *(300/31)
V1 = 300/(5*31) = 1.935
V2 = 300 / (3*31) = 3.225
V3 = 300 /(2*31) = 4.839 (whew, the sum is 10 volts :)
You now have the charge and the voltage on each capacitor
energy stored in each = (1/2) q V in Joules
so just add them up

Answer 2

Whoops, I used ten volts. They said 100
multiply the charges by 10 and the voltages by 10 so the energy by100.