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

Explanation

Let x be the equivalent resisance of infinite network. Sine the net work is infinite, therefore, the addition of one more unit of three resistances each of value of `1 Omega` across the terminals will not alter the total resistance of network i.e., it should remain x.
Therefore the network would appear as shown in the figure and its total resistance should remains x.
There the parallel combination of x and `1 Omega` is in series with two resistors of `1 Omega` each.
The resistance of parallel combination is

`(1)/(R_(P)) = (1)/(X) + (1)/(1) = (1 + x)/(x)`
`R_(P) = (x)/((1 + x))`
`:.` Total resistance of network will be given by
`x = 1 + 1 + (x)/(x + 1) = 2 + (x)/(x + 1)`
`x (x + 1) = 2 (X + 1) + x`
or `x^(2) + x = 2x + 2 ` or `x^(2) - 2 = 0`
Or `x = (2 +- sqrt(4 + 8))/(2) = (2 +- sqrt(12))/(2)`
Total resistance of the circuit shows a full scale defliction for a current of 2.5 mA. How will you convert the meter into
`= (2 +- 2 sqrt(3))/(2) = 1 +- sqrt(3)`
The value of resistance cannot be negative, therefore the resisance of network
`= 1 + sqrt(3) = 1 + 1.73 Omega = 2.73 Omega`
Total resistance of the circuit = 2.73 + 0.5
`= 3.23 Omega`
`:.` Current draw `I = (12)/(3.23) = 3.72` amp