Determine the current drawn from a `12V` supply with internal resistance `0.5Omega`. By the infinite network shown in fig. Each resistor has `1Omega` resistance.

Explanation

Let `x` be the equivalent resistance of infinite net work. Since the net work is infinite, therefore, the addition of one more unit of three resistance each of value of `1Omega` across the terminals will not alter the total resistance of net work, i.e., it should remain `x`.
Therefore, the network would appear as shown in Fig.`2(NCT)`.4 and its total resistance should remain `x`. Here the parallel combination of `x` and `1Omega`is in series with the two resistors of `1Omega`each. The resistance of paralle combination is
`1/R_(p) =1/x+1/1 =(1+x)/x or R_(p) = x/(x+1)`
`:.` Total resistance of net work will be given by `x=1+1+(x)/(x+1) =2+(x)/(x+1) or x(x+1)= 2(x+1)+x or x^(2)+x=2x+2+x or x^(2)-2x-2=0 or x=(2+-sqrt(4+8))/2 =(2+-sqrt(12))/2 =(2+-2sqrt3)/2 =1+-sqrt3`
The value of resistance can not be negative, therefore, the resistance of net work `=1+sqrt3=1+1.73Omega =2.73Omega`
Total resistance of the circuit `=2.73 + 0.5 = 3.23Omega .:` Current drawn, `l=12/3.23 =3.72` amp