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

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,

$\therefore R^{\prime}=2+\frac{R^{\prime}}{\left(R^{\prime}+1\right)}$

$\left(R^{\prime}\right)^{2}-2 R^{\prime}-2=0$

$R^{\prime}=\frac{2 \pm \sqrt{4+8}}{2}$

$=\frac{2 \pm \sqrt{12}}{2}=1 \pm \sqrt{3}$

Negative value of R’ cannot be accepted. Hence, equivalent resistance,

$R^{\prime}=(1+\sqrt{3})=1+1.73=2.73 \Omega$

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, $\frac{12}{3.23}=3.72 \mathrm{~A}$