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

Explanation

The resistance of each resistor connected in the given circuit, `R = 1 Omega`
Equivalent resistance of the given circuit `= R`
The network is infinite. Hence, equilibrium resistance is given by the relation,
`:. R = 2+(R)/((R+1))`
`(R)^(2)-2R -2=0`
`R = (2 pm sqrt(4+8))/(2)=1 pm sqrt(3)`
Negative value of R' cannot be accepted. Hence, equivalent resistance
`R= (1+sqrt(3))=1 + 1.73 = 2.73 Omega`
Internal resistance of the circuit, `r = 0.5 Omega`
Hence, total resistance of the given circuit `= 2.73 + 0.5 = 2.23 Omega`
Supply voltage, `V = 12 V`
According the Ohm's Law, current drawn from the source is given by the ratio, `(12)/(3.23) = 3.72 A`.