Q:

Change in ammeter's indication due to change in resistance

I stumbled on a problem in a textbook, where I was asked to describe the change in ammeter's indication due to change in resistance in circuit: rheostat's slider is moved or key is locked as indicated in the pictures, how ammeter's indication will change? (the generator's resistance is negligible) 
It is clear, that in the first and penult cases the current will go down, so the ammeter's indication will, in  case d) (the first one in second row) the current will go up, so as indicaton, in case c) ("в" in the picture) the first ammeter's indication will go up and the second's down, although it's not clear for me, but the textbook says this is correct, but I was wondering: in the case b) we reduce the resistance and so the current throw the circuit must go up, but in the branches current throw each branch is proportional to resistance of branch, so the the current throw the rheostat must go up. My question is: won't that change in ratio of currents throw branches be compensated with overall current going up? The textbook says so, as the answer to the case b) is that the indication will not change. However, in the case c) we can apply the same argument and conclude that the first ammeter's indication will rise, yes, but the second's one won't change. Althogh this is wrong according to the textbook! It also says, that in the last case the ammeter's indication will not change too. That's what I don't understand at all. Please, help me to get through this!

A:

First circuit: $I_{R_1}=\displaystyle\frac{V}{R_1}$ no matter how many resistors one applies in parallel to $R_1$. The reading of an ammeter in its branch changes only if one changes $V$ and/or $R_1$.
Second circuit: But as soon as one applies resistance in series, it doesn't hold as follows
$$I=\frac{V}{R_A+\displaystyle\frac1{\displaystyle\frac1{R_1}+\displaystyle\frac1{R_2}}}\Rightarrow R_2\uparrow\Rightarrow I\downarrow$$
$$I_{R_1}=\frac{V}{R_A+\displaystyle\frac1{\displaystyle\frac1{R_1}+\displaystyle\frac1{R_2}}}\frac{R_2}{R_1+R_2}=\frac{V}{R_A\left(1+\displaystyle\frac{R_1}{R_2}\right)+R_1}\Rightarrow R_2\uparrow\Rightarrow I_{R_1}\uparrow$$

