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

Solution

Let x be the equivalent resistance of the infinite network.
This network consists of infinite sets of three resistors of 1 Ω, 1 Ω and 1 Ω. Adding one more set across AB to the infinite network will not affect the equivalent resistance.

Resistance between A and B

R_{p} = \frac{x}{x + 1}

(∵ \frac{1}{R} = \frac{1}{x} + \frac{1}{1})

Resistance between P and Q

R_{s} = 1 + \frac{x}{x + 1} + 1 = 2 + \frac{x}{x + 1}

This must be equal to the initial resistance x

∴

x = 2 + \frac{x}{x + 1}

\Rightarrow x + x^{2} = 2 + 2 x + x

\Rightarrow

x^{2} - 2 x - 2 = 0

\Rightarrow

x = \frac{2 \pm \sqrt{4 + 8}}{2} = 1 \pm \sqrt{3}

But,

x \neq 1 - \sqrt{3}

∴

x = (1 + \sqrt{3}) Ω = 2.732 Ω

Current I = \frac{e . m . f .}{total resistance} = \frac{12}{2.732 + 0.5}

I = 3.7 A .

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Current Electricity

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

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