just method .... havnt really solved it ....
R2R/R2+R + R1 = R
solve for R[Rnetwork] .........
Vab = i R .....
i = Vab/R .....
i thru R2 = SAY i1
i1R2 = i2R
i1 + i2 = Vab/R ....
i1R2 = Vab/2 ...............
solve and get R1/R2
consider an infinite ladder network shown in figure voltage is applied between points A and B . If the value of voltage is halved after each section , then the ratio of R1/R2 Will be
a. 1
b.1/2
c.2
d.3
just method .... havnt really solved it ....
R2R/R2+R + R1 = R
solve for R[Rnetwork] .........
Vab = i R .....
i = Vab/R .....
i thru R2 = SAY i1
i1R2 = i2R
i1 + i2 = Vab/R ....
i1R2 = Vab/2 ...............
solve and get R1/R2
The method is very simple ..
this question has something hidden..
the question statement itself says that R(the effective resistance) is equal to R2!!
( why? )
now we know that R= R1+R/2
thus, R1=R/2
R1/R=2
but R=R2
R1/R2=1/2