The fig shows a network consisting of an infinite number of pairs of resistors `R_(1) = 2 Omega and R_(2) = 1 Omega`. Since the network is infinite, removing a pair of `R_(1) and R_(2)` from either end of the network will not make any difference. Using this calculate the equivalent `(R)` across points A and B.

(b) Prove that `I_(n)=(I_(n-1)R_(2))/(R_(2)+R)=(I_(n-1))/(sqrt(3)+2)` ltbr Where `I_(n)` and `I_(n-1)` represent the current through `R_(1)` in `n^(th)` and `(n-1)^(th)` segment respectively (see Fig)

(c) If a 20 V battery is connected across A and B find `I_(10)`

Answer

(a) `R=(1+sqrt(3))Omega`
(c) `(20)/((sqrt(3)-1)(sqrt(3)+2)^(9))`