Find the resistacne `R_(AB)` of the frame made of a thin wire. Assume that the number of successively embedded equilateral triangles (with sides decrasing ot half) tends to infifity Side AB is equal to a, and the resistance per unit length of wire is `lambda.`

Explanation

Let `R_(AB) =x` be equivalent resistance of system between A and B. As the resistance of a conductor is directly proportional to length, the equivalnet resistance between `A_(1)` and `B_(1)` wil be x/2. Therefore, the equivalent circuit becomes as given in fig Effective resistance of 2r and x/2 is

`R_(1) = (2x(x)/(2))/(2r +(x)/(2)) = (2rx)/(4r +x)`
Now `R_(1)` is in series with `AA_(1)` and `BB_(1)`, therefore, their effective
resistance is `R_(2) = R_(1) +2r = (2rx)/(4r +x) +2r`
`R_(2)` is the parallel with 2r(of AB), so the net effective resistance
across AB is
`x=(R_(2)xx2r)/(R_(2)+2r)=(((2rx)/(4r +x)+2r)2r)/(((2rx)/(4r+x)+2r)+2r)`
or `3x^(2) + 4rx - 8r^(2) = 0` or `x = (-4r+-sqrt(16r^(2)+4xx3xx8r^(2)))/(2xx3)`
As x cannot be negative, we have
`x =(-4r+sqrt(16r^(2)+96r^(2)))/(6)=((2sqrt7-2))/(3)r`
But `2r = a lambda or r = (a lambda)/(2)`, so
`x =(2(sqrt7-1))/(3)xx(a)/(2)lambda=0.55 a lambda`