Effective capacitance between A and B in the figure shown is (all capacitances are in F )

Step-by-step solution

The points C and D will be at same potenitals since, 3 6 = 4 8 . Therefore, capacitance of 2 ~F will be unaffected. So, the equivalent circuit can be shown as the effective capacitance in upper arm in series, is given by C₁= 3 6 3+6 = 18 9 =2 ~F The effective capacitance in lower arm in series, is given by C₂= 4 8 4+8 = 32 12 = 8 3 ~F Hence, the resultant capacitance in parallel is given by C=C₁+C₂=2+ 8 3 = 14 3 ~F