A 9 V battery with an internal resistance of 0.5 $$\Omega$$ is connected across an infinite network, as shown in the figure. All ammeters $$A_1$$, $$A_2$$, $$A_3$$ and voltmeter $$V$$ are ideal. Choose the correct statement.

Let the total resistance of the network be R.

Since the network is infinite, the resistance of the network connected in parallel to the branch with ammeter $$A_2$$ is also R.

Thus, $$R=1+1+\frac{4\cdot R}{4+R}$$

$$R=2+\frac{4R}{4+R}$$

$$4R+R^2=8+6R$$

$$R^2-2R-8=0$$

$$\left(R-4\right)\left(R+2\right)=0$$

$$\therefore\ R=4Ω$$ 

Current passing through ammeter $$A_1$$ = Total current in the circuit = $$\frac{E}{R+0.5}$$

$$\therefore\ I_{A_1}=\frac{9}{4+0.5}=2A$$

Reading of $$V$$, $$V=E-I\cdot0.5=9-2\cdot0.5=8V$$