Question

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

• A. $3.7A$
• B. $4A$
• C. $3A$
• D. $2A$

Answer 1

The correct option is A $3.7A$

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

This network consists of the infinite sets of $3$ resistors $1\Omega ,1\Omega ,1\Omega$.

Adding one more set across $AB$ to the infinite network will not affect the equivalent resistance.

A resistance between $A$ and $B$ is,

$R_P=\frac{X}{X+1}(\because \frac{1}{R}=\frac{1}{X}+\frac{1}{1})$

Resistance between $P$ and $Q$

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

This makes be equal to the initial resistance $X$.

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

$⟹X+X^2=2+2X+X$

$⟹X=\frac{2\pm \sqrt{4+8}}{2}=1\pm \sqrt{3}$

But $X\ne 1-\sqrt{3}$ [$\because$ Resistance cannot be negative]

$\therefore X=(1+\sqrt{3})\Omega =2.732\Omega$

$I=\frac{emf}{Totalresistance}=\frac{12}{2.732+0.5}$

$=3.7A$