Question

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.

Answer 1

Here,

Let $X$ be the equivalent resistance of the infinite network

This network consists of infinite sets of three resistors of $1\Omega$,$1\Omega$ and adding one more set across $AB$ to he infinite network will not affect the equivalent resistance.

We know that

Resistance between $A$ and $B$

$R_p=\frac{x}{x+1}$ $(\therefore \frac{1}{R}=\frac{1}{x}+\frac{1}{1})$

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

This must be equal to the initial resistance $x$

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

$\Rightarrow x+x^2=2+2x+x$

$\Rightarrow x^2-2x-2=0$

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

But $x\ne 1-\sqrt{3}$

$\therefore x=(1+\sqrt{3})\Omega$

$=2.732\Omega$

Current $I=\frac{e.m.f}{Totalresistance}=\frac{12}{2.732+0.5}$

$I=3.7A$