The problem with just letting the equation define it is what happens when you get to the node on the other side of that component and now write I10 as the current leaving that node? You now have I10 defined as going in both directions. It HAS to go into one node and out of the other, so it needs to be indicated which it is.Sorry for confusion, I thought it is a common thing...I mean I learned from some on-line course that when doing nodal analysis with KCL you start by default, defining all currents as coming out of the node in question. Now, if some of the currents do have direction already defined then you reflect it accordingly but those that don't you define as the coming of the node current. I understand that there are obviously more ways of doing it. I have to see how to inject the diagram images in here so I could make it more visual, apologize again for confusing the matters here.
Let's take it step by step.Still confusing.....just looked again....since the direction for Ix is defined in assignment it still has to be into the node but with negative value. Then Isource defined as well into the node...but with positive value (since it is a source?) and Iab then has to have positive value and out of the node. All that is when we have Vab connected to A & B terminals with + to A and - to B ....is it better picture now ?
If Vab=30V, then the A node is 30V greater than the B node. The very notation Vab means "the voltage at Node A minus the voltage at node B" (assuming we are case-insensitive).@screen1998....the thing is that I don't get 6A as you say I would. That is because if I apply Vab=30V (+B/-A) the currents are as follows Ix=1A, 2Ix=-2A, I10=-1A (Ix+2Ix=-1)
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