Hi,
One way to calculate the theoretical readings of ideal ammeters placed at those locations is to apply Kirchhoff's laws to the nodes and loops of the given circuit.
In analyzing the circuit using Kirchhoff's Current Law, we would write down the sum of all incoming and outgoing branch currents at each node of the circuit.
Note that the given illustration of the circuit has 2 extraneous branches: the 2 branches that have no circuit elements have no effect on the branch currents or branch voltages in the rest of the circuit, assuming those branches represent ideal conductors. As such, it may be best to redraw the diagram, reducing those extraneous branches to nodes (see the left diagram in the attached image).
The redrawn circuit diagram has only 4 nodes, labeled a, b, c, and d in the attached diagram (the blue dots in the right diagram).
We may then set up the directions of branch currents into and out of each node by following the associated variables convention, or passive sign convention (if the actual current direction is opposite to this, we will just get a negative branch current value in the solution).
There are 7 branches in total, so we have 7 branch currents to find. We will let current flowing into a node be added to the sum, while current flowing out is subtracted (the reverse can also be done; the results will be the same). For the node labeled "a", for example, we get the equation:
i₁ + i₃ - i₄ - i₅ = 0
Similarly, for nodes b and c, we get:
i₄ + i₅ - i₂ = 0
i₆ + i₇ - i₃ = 0
There is no point in writing down the current law for node d because it will be linearly dependent on the equations we have already written, and as such carries no new information.
So far we have 3 linearly independent linear equations in 7 unknowns, so from linear algebra, we know that we need at least 4 more linearly independent equations in order to find a unique solution.
We can get those equations from applying Kirchhoff's Voltage Law to the 4 internal loops of our circuit diagram. Recall Kirchhoff's Voltage Law states that the sum of the branch voltages in a circuit loop must be 0.
Going clockwise around each loop, we get the equations:
60 - 4i₃ - 9i₆ - 2i₂ - 6i₄ = 0
9i₆ - 18i₇ = 0
5i₁ + 3i₅ + 2i₂ - 25 = 0
6i₄ - 3i₅ = 0
Now that we have an inhomogeneous system of 7 linearly independent linear equations in 7 unknowns, we know from linear algebra that there is either 1 unique solution or no solution.
In this case, there is 1 unique solution, which gives us the branch currents for the associated ammeters: i₁ = 1 A, i₂ = 5 A, and i₃ = 4 A.
Note that you can also reduce your work by treating the pair of 3Ω and 6Ω resistors as a single 2Ω effective resistor, and similarly treating the pair of 9Ω and 18Ω resistors as a single 6Ω effective resistor, since there are no ammeters on those separate branches.
Note that if you have already derived the circuit laws for combining resistors in series and in parallel from Kirchhoff's circuit laws, then you can simplify the diagram considerably, first by combining the two pairs of parallel resistors into effective resistors, then by combining these effective resistors with the resistors they are in series connection with (see the attached diagram), yielding an effective circuit diagram that differs considerably from the original diagram, but has the same branch currents through the three given ammeter locations.
View attachment CircuitDiagram2-01.png
This new diagram has only two nodes, and thus only one linearly independent Kirchhoff's current law:
i₁ + i₃ - i₂ = 0
The remaining two linearly independent equations that we need can come from applying Kirchhoff's voltage law to the two internal loops:
4i₂ + 10i₃ - 60 = 0
25 - 5i₁ - 4i₂ = 0
This inhomogeneous system of 3 linearly independent equations in 3 unknowns has the same unique solution as before.