Given : $E = 2 V, r = 1.2 \Omega, R =$ (external resistance) Let $4.5 \Omega$ and $9 \Omega$ connected in parallel, then equivalent resistance $\; \;\frac{1}{R_p} = \;\frac{1}{4 \cdot 5} + \;\frac{1}{9}$ $\; \;\frac{1}{R_p} = \;\frac{2+1}{9} = \;\frac{3}{9} = \;\frac{1}{3}$ $\; R_p = 3 \Omega$ Now $0.8 \Omega$ and $R_p$ resistance in series, then total resistance $R = 3 + 0.8 + 1 \cdot 2 \Omega = 5 \Omega$ Current in the ammeter $I \; = \; \frac{ \; \text{ Total e.m.f. } \; }{ \; \text{ Total resistance } \; }$ $\; = \; \frac{2}{5} = 0.4 A$