What is the effective capacitance of this circuit?

Capacitors in parallel add. 26+27 = 53 μF on each side of the circuit. Capacitors in series combine like resistors in parallel. Two equal capacitors in series will have an equivalent capacitance of half. 53/2 = 26.5 μF is the capacitance of the circuit.

Only half the voltage drop will occur over the parallel combination on the right. The energy stored in a capacitor is given by

1/2 CV^2 = 1/2 (27 μF)(32 V)^2 = 13.8 mJ

C1 = 26 μF , C2 = 27 μF , C3 = 26 μF , C4 = 27 μF

C1 and C2 are connected in parallel and C3 and C4

are connected in parallel.

Left hand combination and the Right hand combination

are in series.

Left hand side : Cx = C1 + C2 = (26 + 27) μF = 53 μF

Similarly, Right hand side :

Cy = C3 + C4 = (26 + 27) μF = 53 μF

1/ Ceff = 1/Cx + 1/Cy = 2/53

=> Ceff = 53/2 μF = 26.5 μF

Vx = Vy = 64/2 = 32 V

Energy stored in the C4 = 27 μF capacitor

= E = (1/2)C4 (Vy)² = (1/2)(27)(32)² μJ

= 13824 μJ = 13.824 mJ