Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

The resistance of each resistor connected in the given circuit,

R = 1 Ω

Equivalent resistance of the given circuit = R’

The network is infinite.

Hence, equivalent resistance is given by the relation,

R’ = 2 + R’/(R+1)

(R’)² -2 R’ -2 = 0

R’ = [2 ± √ (4+8)]/2

= [2 ± √12]/2 = 1 ± √3

Negative value of R’ cannot be accepted.

Hence, equivalent resistance,

R’=(1 + √3)= 1 + 1.73 = 2.73 Ω

Internal resistance of the circuit, r = 0.5 Ω

Hence, total resistance of the given circuit

= 2.73 + 0.5 = 3.23 Ω

Supply voltage, V = 12 V

According to Ohm’s Law, current drawn from the source is given by the ratio, 12/3.23= 3.72 A