shashi K Sharma
Last Activity: 10 Years ago
Dar Ashwin
The case illustrated is a case of infinite series whose each repeating step is double the prior step. Let the Req=r (let) then the equivalent resistance after the first sep may me written as 2r. Redraw the cir cuit then we get that the vertical resistance R is in parallel with 2r. Thus the equivalent resistance between will be equal to 2R + (2rR)/(2r+R) = r (as asuumed earler that equivalent resistance between A & B is r).
This leads to a quadratic equation whose solution comes as (5 +root41)R/4