a
The given circuit can be redrawn as,

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 \(A_{1}=\frac{V}{(R+r)}\)

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