How can we prove mathematically that these measured currents are correct? Why is it that one of the currents is the sum of the other two currents?
With three batteries, three resistors and two nodes, this looks kind of ugly. How do we know that the current is 0.61 amps at that location? Well, coming to that conclusion is not exactly forthcoming. You must write down what you know for sure, and then manipulate that data to eventually find a number. It sounds pretty terrible, but it's really not that bad.
Let's choose a node to calculate the current through each of the wires. We can do this by writing down what we can tell about the circuit pictorially. After we have chosen the node, we have three "loops" to choose from, but only two can be considered loops under Kirchoff's definition. In other words, only two of the three "loops" above are part of the circuit. That is to say that from either node, two of the wires are in parallel and the third wire is in series with the two parallel wire combination. This can be explained in several ways, but basically it stems from Kirchoff's 2 Rules; Kirchoff's Junction Rule and Loop Rule, which are based on the law of conservation of charge and the law of conservation of energy, respectively. These rules state that the amount of charge/electrons going into a junction/node equals the amount of charge/electrons going out [conservation of charge], and that the sum of all of the voltages in a loop is equal to 0 [conservation of energy].
So which wire(s) draw current from the node and which wire(s) donate current to the node? To answer this question we must consider each potential loop, of which there are three. Consider the possible loop with the 6V and 8V batteries at odds with one another. The 8V battery will win, and the current will flow as if there are 2V applied to it in the direction of the 8V battery. Now consider the loop between the 8V and 12V batteries. The 12V battery wins. Finally consider the loop between the 12V and 6V batteries. Again the 12V battery wins. Since the 12V battery wins over each of the other batteries and its voltage is applied to each of the other wires equally at 12V each, the current will flow in the direction that the 12V battery is oriented in, from positive to negative, of course.
So now that we know what our loops are, we can use Kirchoff's Junction Rule to state the relation between the currents flowing in and out of the node we have chosen:
OK, so we have 3 unknowns and 1 equation. We need two more equations. We will obtain these using Kirchoff's Loop Rule. These equations will relate the Voltages across each of the two loops.
Now we are left with a slightly tricky algebra problem. But do not despair. We can figure it out! Let's come up with a game plan based on the equations that we have. We will be able to figure out the current I1 if we manipulate the 2nd and 3rd equations to solve for I2 and I3 in terms of I1:
