Two capacitors of capacities c1=3uf and c2=6uf arranged in series are connected in a parallel with third capacitor c3=4uf. The arrangement is connected to a 6.0 V battery. Calculate the total energy stored in the capacitors.

Answer 1

C = C1*C2/(C1+C2) = 3*6/(3+6) = 2 uF.

Ct = C + C3 = 2 + 4 = 6 uF.

Total Energy = 0.5C*E^2 = 0.5*6*6^2 = 108 J.

Answer 2

In the numerical value, C is value is 6 uf, otherwise it's OK

Answer 3

All was correct except si unit to micro joule