# Current through a short circuit in the Wheatstone Bridge configuration

The task was to replace part of a circuit with corresponding Norton's generator. To calculate the current of the generator, I had to calculate the current going through the short circuited nodes 1 and 2, pictured below: There are a couple of things I do not understand.

First, let's look at the node 3. Current I is going in, and currents through R4 and R1 are going out. Then, those two currents are going in nodes 1 and 2, respectively. But if there is a short circuit between 1 and 2, currents have the option to go through the path with the least resistance, short circuit, and avoid R5 and R3 completely. In that case, current through the short circuit will be the difference between them, since they are of opposite direction. Current through R4 and R1 can be calculated by considering a current divider, that is, by combining Kirchoff's Laws. However, according to my textbook, this would not be the solution to the problem.

Another thing that I don't understand is, if currents don't go through R5 and R3 at all, then they must come back where they came from, node 3, but this current would colide with one of the ideal generator. I have a feeling that the ideal current source pushes them to go to R5 and R3 as well and come through the other side, node 4.

My question is: What is the current through short circuited nodes 1 and 2?