An infinite ladder is constructed with `1(Omega)`and `2(Omega)`resistor as shown in figure.(a)Find the effective resistance between the point A and B. (b) Find the current that passes through the `(2 Omega)` resistor nearest to the battery.

Explanation

(i) `2Omega` (ii) 1.5 A
(i) Let `R_(AB)=x` Then we can break one chain and connect a resistance of magnitude in place of it.Thus, the circuit remains as shown in figure Now, are in parallel. So, their combined resistance is `(2x)/(2+x) or R_(AB)=1+(2x)/(2+x)`
But `R_(AB)` is assumed as x. Therefore `x=1+(2x)/(2+x)`
Solving this equation ,we get `x=2Omega`

(ii) Net resistance of circuit `R=1+(2 times 2)/(2+2)=2 Omega`
`therefore` Current through battery `i=6/2=3A`
The current is equally distributed in `2Omega and 2Omega` resistances , Therefore the desired current is `i/2 or 1.5A`.