# Currents through a circuit (junctions/series/parallel)

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1. Sep 7, 2015

### RelativeJosef

1. The problem statement, all variables and given/known data

Find currents through M and N

2. Relevant equations
Current out must be current in.

3. The attempt at a solution
I think that it is simply 3 parallels running with 5A of current. So, M is in series with the 2A current thus also has 2A of current. Would N also be in series with the two elements with 1A of current, thus making it's current also 1A?

I am confused because of the third pathway between the second and third parallel. Would the current try to skip the path with two elements in the bottom right?

So I think I am certain M has 2A of current. Then the question remains, Does N have 2A of current, or does the current run through all three elements in the third parallel making it's current 1A?

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2. Sep 7, 2015

### phyzguy

Your answer is incorrect. Try thinking of what happens at each junction, where the current can take multiple paths. At each of these points, the current out has to equal the current in. So there is 5A coming out of the top of the battery. 1A goes down through the square, so how much is left to travel through the wire between the top of the square square and the top circle? Of this, 2A flows through the top circle. So how much is left after the 2A flows through the top circle? All of this has to flow through N. Similarly, you can work backward from the bottom of the battery. 5A is flowing into the bottom of the battery. 1A is flowing in from the square, so how much is flowing through the wire between the bottom of M and the bottom of the square? You know 1A is coming from the bottom circle, so how much is coming through M?

3. Sep 7, 2015

### RelativeJosef

There are 4A remaining

2A remain

So N must have 2A, but I am still confused about M. I would think it to have 2A as well? Working backwards from the battery was something I did not think about. But since 1A comes back to the battery from the first square, an unknown amount leaves M, and then 1A comes back from the final branch. This means M would have 3A? But how would the square above M have 2A and then it suddenly gains current to become 3A?

Thank you for your help, any more would be greatly appreciated.

4. Sep 8, 2015

### phyzguy

M has 3A like you said. It gains current because there is 1A flowing through the wire connecting N to M, and this 1A adds to the 2A flowing through the top circle to bring the current through M up to 3A. The lower right circles both have 1A.