You are given thirty two capacitors each having capacity 4 μF. How do you connect all of them to prepare a composite capacity having capacitance 8 μF?

Let m condensers are joined in series and n such groups are in parallel.Capacity of m condensers connected in series1C′=1C1+1C2+…=mC or C′=Cm=4mWhen n such groups are connected in parallel, thenCeff=nC′=4nmGiven that Geff=8, ∴8=4nmOr     nm=2  or  n=2m                         ...(1)Further                 mn=32                           ....(2)Now    m×2m=32  or  m=4and     n = 2 x 4 = 8