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Find eq resistance b/w the terminals a and b of the following infinite ladder networks .

My try :

In this I am unable to find $R_{bc}$

Why are you not able to find $R_{BC}$?
I do not know the sum of $1+ \frac{1}{2} + \frac{1}{3} +\frac{1}{4} .........$
It is called  the harmonic series.

1 vote

As you have realised, all points along the wire running between the 2 infinite ladder networks are at the same potential, so the 2 networks can be separated. The circuit then consists of the 2 networks $$R_{eq}=R_1+R_2$$
Network $R_2$ consists of an infinite series of resistors in parallel, so its equivalent resistance is given by $$\frac{1}{R_2}=\frac{1}{R}+\frac{1}{2R}+\frac{1}{3R}+\frac{1}{4R}...=\frac{1}{R}(\frac11+\frac12+\frac13+\frac14+..)$$ The series in brackets is the harmonic series. Its sum diverges to infinity, so $R_2 = 0$.
You have also realised that network $R_1$ can be solved by symmetry.
Removing the 2 leading resistors $R$ leaves a similar infinite ladder with everything scaled down by a factor of $\frac12$ - ie with $R$ replaced by $\frac12 R$. We can assume that the network is linear, so if the leading resistors have value $kR$ then the equivalent resistance will be $k$ times that when the leading resistors have value $R$ : $$R_{eq}(kR)=kR_{eq}(R)$$ $$R_{eq}(\frac12 R)=\frac12 R_{eq}$$
Therefore $$R_1=R+\frac{\frac12 R_1 R}{\frac12 R_1+R}$$ which has the solution $R_1=\sqrt2 R$.