as $4\, \Omega$ and $x \,\Omega$ are parallel

$x^{\prime}=\frac{1}{4}+\frac{1}{x}=\frac{(4+x)}{4 x} \quad x^{\prime}=\frac{4 x}{4+x}$

and $ 1 \,\Omega$ and $1\, \Omega$ are also in series $x^{\prime \prime}=2\, \Omega$ Now equivalent resistance of circuit

$x=\frac{4 x}{4+x}+2=\frac{8+6 x}{4+x}$

$4 x+x^{2}=8+6 x$

$x^{2}-2 x-8=0$

$x=\frac{2 \pm \sqrt{4-4(1)(-8)}}{2}=\frac{2 \pm \sqrt{36}}{2}$

$=\frac{2 \pm 6}{2}=4\, \Omega$

Reading of Ammeter $A_{1}=\frac{V}{(R+r)}$

$A_{1}=\frac{9}{4+0.5}=2$ $Ampere$