Find the equivalent capacitance between the terminals A and B in the given figure. Given C=1F.

Explanation

Let C. = equivalent capacitance between A and B.
Since the network is infinite, so if-we move one set of capacitors from the chain, even then the network will remain infinite.
It means, the capacitance between D and E can be taken as C (Figure (Q)

The effective capacitance of parallel combination of C and C. as shown in figure is C.. = C + C..
Now C and C" are in series, so equivalent capacitance of the network is given by:
`1/C^(.) =1/C + 1/C^(.) =1/C + 1/(C +C.)`
`=(C. +2 C)/(C(C+C.))`
or `C.^(2) + C C. -C^(2)=0`
or `(-C +- sqrt(5C^(2)))/2`
Since capacitance cannot be negative,
`therefore C. = (-C + sqrt(5)C)/2 =C/2 [sqrt(5)-1]`
`=C/2 [2.24-1]=C/2 xx 1.24`
But `C = 1muF` (given)
`therefore C. =(1.24)/2 = 0.62 muF`