Fig. shows an infinite circuit formed by the repetition of the same link, consisting of resistance `R_(1) = 4.0 Omega` and `R_(2) = 3.0 Omega`. Find the resistance of this circuit between points `A` and `B`.

Explanation

Let `R_(0)` be the resitance of the network,

then, `R_(0) = (R_(0) R_(2))/(R_(0) + R_(2))` or `R_(0)^(2) - R_(0) R_(1) - R_(1) R_(2) = 0`
On solving we get,
`R_(0) = (R_(1))/(2) (1 + sqrt(1 + 4 (R_(2))/(R_(1)))) = 6 Omega`