  Question

# If each capacitor is of capacitance  4 μF. The equivalent cpacitance (in μ F) between terminals A & C  is x. Then the value of  4x is . Solution

## For above arrangements following points can be noted: (1) The charge recombines at C identical to the way it splits at A. (2) Potentials at E and F are same. So, capacitor in between them can be replaced by conducting wire. (3) Potentials at H and G are same. So, capacitor in between them can be replaced by conducting wire. (4) Connections at B and D can be removed. Above circuit can be rearranged as On simplification we get As capacitors of capicatance C and C2 are in parallel (circled in above figure) effective capacitance due to them is  C+C2=3C2 Now capacitors of capacitance C, C, 3C2 are in series. So, effective capacitance due to them is 113C/2+1C+1C =113C/2+2C =3C2×C23C2+C2=3C242C=3C8 So, circuit can be further simplified as From above circuit , effective capacitance between A and C is  CAC=3C8+3C8+C=7C4 Given C=4 μF x=7×44 x=7 4x=28  Suggest corrections   