Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in the figure. Each resistor has 1 Ω resistance.
Solution
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 ± sqrt(4 + 8))/2`
= `(2 ± sqrt12)/2`
= `1 ± sqrt3`
Negative value of R’ cannot be accepted. Hence, equivalent resistance,
R' = `(1 + sqrt3)` = 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
