(a) Find the equivalent resistance between A and B of the network extending off to the infinity shown in the figure .

(b) If `E = 12 V , R_(1) = 1 Omega and R_(2) = 4 Omega`. Find the current in `R_(2)` nearest of the battery .

Explanation

(a) Let the equivalent resistance of the given infinite ladder is R. Since the network extends to infinity , the resistance of the network to the right of the points C and D is also equal to R . Hence

or ` R = 2R_(1) +( R R_(2))/(R + R_(2)) = (2R_(1)R_(2) + 2R_(1)R_(1) + R_(2)R)/(R + R_(2))`
or `R^(2) + R_(2)R = 2R_(1) R_(2) + 2R_(1)R + R_(2)R`
or `R^(2) - 2R_(1)R - 2R_(1)R_(2) = 0`
or `R = 2R_(1) pm sqrt(4R_(1)^(2) + 8R_(1)R_(2))/(2) = R_(1) + sqrt(R_(1)^(2) + 2R_(1)R_(2))`
(Negative resistance is not acceptable )
(b) Where `R_(1) = 1 Omega , R_(2) = 4 Omega, ` the equivalent resistance
`R= R_(1) + sqrt(R_(1)^(2) + 2R_(1)R_(2)) = 4 Omega`
`therefore ` The net currents is ` i = (12 V)/(4 Omega) = 3A`
So, the current in each of `4 Omega` resistor in figure (c) is `(i)/(2) = 1.5A`
That is , the current in `R_(2)` nearest to the battery is 1.5A