A circuit contains an ideal battery, three resistors, and two ideal ammeters. The ammeters read 0.2 A and 0.3 A. After two of the resistors are switched, the readings of the ammeters did not change. Find the battery current  

As the ammeters are ideal, we can conclude that the three resistors $R_1,R_{2}andR$ are actually connected in parallel, the equivalent resistance being

$R_{eq}=\frac{R_1{R}_2{R}_3}{R_1{R}_2+R_1{R}_3+R_2{R}_3}$ In the figure, the electric currents on each resistor and battery are shown. 

The readings of the two ammeters are different and, consequently, by simple symmetry considerations, resistors  $R_{1}and{R}_3$ are necessarily different. In other words, if we interchange  $R_{1}and{R}_3,$ the readings of the ammeters will also be interchanged. Therefore, we have two possible alternatives. 

$R_2=R_1$ (if resistors)  $R_{2}and{R}_1$ switched, the readings of the ammeters do not vary) or  $R_2=R_3(R_{2}and{R}_3$ can be switched and the ammeters do not change). The two solutions are:

If $R_2=R_1$ we obtain $I_1=I_2=0.15Aand{I}_3=0.05A,$ and the battery current is I = 0.35 A

If $R_2=R_3$ we obtain  $I_2=I_3=0.1Aand{I}_1=0.2A,$ and the battery current is I= 0.4 A.