A 9   V battery with an internal resistance of 0.5  Ω is connected across an infinite network…

In the given infinite network problems, let, $X$ be an equivalent resistance. If we connect one section of an infinite network with the replacing network using an equivalent resistance, then in this new equivalent circuit the equivalent resistance will still remain the same.

Let, $R_{\mathrm{eq}}=x$
$x=\frac{4x}{4+x}+2$
$x=\frac{8+6x}{4+x}$
$4x+x^2=8+6x$
$x^2-2x-8=0$. 

On solving,

$x = 4 , x= -2$
Here, the negative value of resistance is not possible.

So, resistance is $x=4 \mathrm{\Omega }$ 

Reading of $A_1$ is $I=\frac{V}{X+r} ...\left(1\right)$, 

where, $V=9 \mathrm{Volt}, r = 0·5 \mathrm{\Omega } , R=4 \mathrm{\Omega }$. 

Then, from the above equation,

$I =\frac{9}{4+0.5}=2 \mathrm{A}$.