Question

Two parallel plate capacitors A and B having capacitance 1µF and 5 µF are charged separately to the same potential 100V. They are then connected such that +ve plate of A is connected to –ve plate of B. Find the charge on each capacitor and total loss of energy in the capacitors

Answer 1

The capacitors are connected in series. 

Energy U1 + U2 = 1/2 C1 V1²
   = 1/2 * 10⁻⁶ * 100² + 1/2 * 5 * 10⁻⁶ * 100²
   = 0.03  Joules 

The total charge gets redistributed. So there is work done.  That becomes the loss.

Effective capacitance of both =  C1 C2 / (C1 + C2) = 5/6 μF
Total voltage across both = 100 + 100 = 200 V
Final energy = 1/2 * 5/6 * 10⁻⁶ * 200² J = 0.0167 J

Loss = 0.0133 J