Question

An infinite ladder network of resistances is constructed with $1\Omega$ and $2\Omega$ resistances, as shown in figure.
The $6V$ battery between $A$ and $B$ has negligible internal resistance.
(a) Show that the effective resistance between $A$ and $B$ is $2\Omega .$
(b) What is the current that passes through the $2\Omega$ resistance nearest to the battery?

Answer 1

(a) Let $R_{AB}=x.$ Then, we can break one chain and connect a resistance of magnitude $x$ in place of it.
Thus, the circuit remains as shown in figure
Now, $2x$ and $x$ are in parallel. So, their combined
resistance is $\frac{2x}{2+x}$ or $R_{AB}=1+\frac{2x}{2+x}$
But $R_{AB}$ is assumed as $x$.
Therefore $x=1+\frac{2x}{2+x}$
Solving this equation, we get
$x=2\Omega$ Hence Proved.
(b) Net resistance of circuit
$R=1+\frac{2\times 2}{2+2}=2\Omega$
$\therefore$ Current through battery $i=\frac{6}{3}=3A$
This current is equally distributed in $2\Omega$ and $2\Omega$
resistances. Therefore, the desired current is $\frac{i}{2}$ or $1.5A$.