# Nodal Equations Confusion????

1. Sep 4, 2016

### Marcin H

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

2. Relevant equations
Node Voltage Method
V=IR

3. The attempt at a solution
So I am very confused about the node voltage method. For this example, the first equation makes sense to me. Current in equals current out. 6A in and then the rest makes sense. Is this how you do node voltage method? Just look at the currents going in and out? And then forming equations? If so the second equation doesn't make sense to me. I was told by another student who took circuit analysis that the second equation I wrote in the picture below is correct. I don't understand why. It doesn't follow the current in = current out logic. I was also told that current leaving the node is positive and entering is negative, but that logic doesn't seem to agree with the "correct" equation. With that logic I would get:

(V2-0)/2 + 2A - (V2-V1)/3 = 0

Why is that wrong?

Solution: EDIT* Using the equations in the picture, I get values of V1=16V and V2=4V. So I get nice values, but I am still not sure why this works if this is the correct method. Sorry for the multiple edits. I kept messing up my matrix

Last edited: Sep 4, 2016
2. Sep 4, 2016

### axmls

Don't get caught up in what's supposed to be positive and what's supposed to be negative here. The core of the nodal equations is that the total current entering the node is the same as the total current exiting a node. In this case, you clearly have 2 A leaving the node because of the current source, but if you want the current from the current source to be the current that is entering the node, you must make that negative to change its direction. The other two currents are straightforward as in the first equation.

I assume you meant the denominator to be 3 in the third term here. So it starts off well. You've got that $V_2/2$ and that's leaving the node. You've got the 2 A, which is also leaving the node. Now move the third term on the other side. Is $(V_2 - V_1)/3$ the current entering the node?

3. Sep 4, 2016

### Marcin H

The way I wrote it: (V2-V1)/3 would mean that the current is leaving the node. Which is wrong right? Current is entering the node there. Should it be (V1-V2)/3 and then add it to the other side? So:

(V2-0)/2 + 2A = (V1-V2)/3

Is that correct?

4. Sep 4, 2016

### axmls

Yes, that is correct, but I must emphasize here: don't get overly concerned with where the current is really entering or leaving. You can choose to treat any path as the current leaving or the current entering. That is, you can treat the current source, for example, as either 2 A leaving or -2 A entering. If you chose wrong, the current you get at the end will be negative, but that's alright.

Personally, I think the best way to do it is to write it as "the sum of all the currents leaving a node is 0." That makes sense, of course, there's not going to be any net current produced in a node--every bit of current that enters will also leave, so there is no net current leaving. That way, you don't have to worry about what's going in and what's going out: simply write every current as leaving the node, add it all up, and set it equal to 0. It's the same thing, but you don't get caught up in the question of what's entering and what's leaving.

5. Sep 4, 2016

### Marcin H

Ok, that clears it up more. I will use that method from now on then. Do you know what I could say for part C? I'm not too sure how the answer relate there.
In Part A I got V1=16V and V2=4V
In Part B I got V0=-16V and V2=12V

i understand why the V2's are different. Because the grounds change you are looking at the potential across different points in the circuit. But what is the reason for V1=-VO? Assuming I did this correctly, what does that mean exactly? Does this show the correct direction of current flow through the circuit? Does this mean that putting the ground at V1 will give you incorrect voltages? What I get from this is that current is definitely flowing down towards V0.

6. Sep 4, 2016

### axmls

Well, in part B, the polarity between V0 and V1 is just flipped, isn't it?

When you actually calculate the currents, the "correct direction" is simply the direction that makes their flow positive. That doesn't mean a negative answer is wrong.

7. Sep 4, 2016

### Marcin H

Yes it is. So my answer makes sense then. Ok cool.

[/QUOTE]When you actually calculate the currents, the "correct direction" is simply the direction that makes their flow positive. That doesn't mean a negative answer is wrong.[/QUOTE]
Ok so does that mean that V1 is not or should not be used as ground or something? Does this mean that V0 is the correct node to have a ground at? I'm confused about what this question is getting at.

8. Sep 4, 2016

### axmls

The biggest thing to take away is that voltage is always measured relative to some reference, so a voltage by itself means nothing. You can generally choose what that reference is. As far as actual grounds in real life circuits, that will make more sense once you've gotten farther into your curriculum.