Nodal analysis of a fairly simple circuit

[Feodalherren]

Homework Statement

Given this circuit, find i.

Homework Equations

Nodal analysis, Ohm's law other basic EE equations.

The Attempt at a Solution

The stuff in blue is given in the problem. The red is my attempt at a solution.

So first off a general question that I have is, when I'm doing nodal analysis can I always say that all of the currents are going out of the node and thus treat them as positives? For an example, in this case I'd say that the current from the V1 node going toward V2 is positive when I'm analyzing node one. But when I'm analyzing node 2, can I say that the same current is going toward V1 or do I have to stay consistent and say that the current is going IN to node 2 at that point?

Here's my attempt, node 1:

(60-V1)/2 + (V2-V1)/8 + V1/10 = 0

node 2:
(V1-V2)/8 + (60-V2)/10 + (3/2)(V1-60) = 0

Simplify and solve the system,
I found
V1 = 44.12 V
V2 = -54.71 V (this makes me question my result, is this a reasonable answer?)

I = V/R = 44.12/2 = 7.94 A.

Incorrect answer.

 

[gneill]

Homework Statement

Given this circuit, find i.

Homework Equations

Nodal analysis, Ohm's law other basic EE equations.

The Attempt at a Solution

The stuff in blue is given in the problem. The red is my attempt at a solution.

So first off a general question that I have is, when I'm doing nodal analysis can I always say that all of the currents are going out of the node and thus treat them as positives? For an example, in this case I'd say that the current from the V1 node going toward V2 is positive when I'm analyzing node one. But when I'm analyzing node 2, can I say that the same current is going toward V1 or do I have to stay consistent and say that the current is going IN to node 2 at that point?

You can do it either way provided that you are consistent when writing the equation for an individual node. Consider that changing the direction of the all the currents for a given node is equivalent mathematically to multiplying both sides of the equation by -1, which still leaves you with the same equation.

Here's my attempt, node 1:

(60-V1)/2 + (V2-V1)/8 + V1/10 = 0

Uh oh. Problem with the last term. You're summing currents entering the node but that last term is a current leaving the node. You would have to write (0 - V1)/10 for the current to be entering the node. (This is why my personal preference is to always sum currents leaving the node. Resistors from a node to ground are a very common occurrence!).

node 2:
(V1-V2)/8 + (60-V2)/10 + (3/2)(V1-60) = 0

Again the last term is a problem. i is shown flowing from the 60V source to node V1. So that current must be (60 - V1)/2. You have it the other way around (flowing from V1 to 60V).

Make the corrections and have another go at solving the equations.

 

[Feodalherren]

I'm still confused. Can't I assume that ALL currents at ALL nodes are going OUT of the node? I think what you're saying is that once I pick a direction for one node, I have to stick with it throughout the problem. So for instance, on the top-left side if I pick i0 as leaving the 60V node I have to pick it as going into the node at V1. If I pick it as leaving node V1 my result will be incorrect. Did I understand you correctly?

Reworked the problem with the corrections and got the right answer, thanks!

 

[gneill]

I'm still confused. Can't I assume that ALL currents at ALL nodes are going OUT of the node?

Yes.

I think what you're saying is that once I pick a direction for one node, I have to stick with it throughout the problem.

No, I'm saying that once you pick a direction for a given node you have to stick to that choice for all the currents for that node.

So for instance, on the top-left side if I pick i0 as leaving the 60V node I have to pick it as going into the node at V1. If I pick it as leaving node V1 my result will be incorrect. Did I understand you correctly?

If you're writing the node equation for V1 then you have a choice of summing currents coming in, or going out.

If you're summing currents coming into V1 then for that branch you would write the term:

(60 - V1)/2

If you're summing currents leaving then it would be:

(V1 - 60)/2

If it's the defined current i that you need to for substituting into the expression for the controlled current source then you must heed its defined direction on the diagram. It would be i = (60 - V1)/2. But this does not interfere with your writing of branch terms for the node; It's a separate equation altogether.

Reworked the problem with the corrections and got the right answer, thanks!

Excellent. Glad to hear it.

 

[Feodalherren]

Sorry but I still don't understand.

You said I CAN assume all currents at all nodes are exiting but then you said that I have to heed the direction of the given i.

Maybe I'm phrasing my question in a strange way, let me try the math. Can I do this:

Node 1:

(60 - V1)/2 + (V2-V1)/8 + (0-V1)/10 = 0

Node 2:

(60 - V2)/10 +(V1-V2)/8 + (0-3i) = 0

Would this be correct?

 

[gneill]

Sorry but I still don't understand.

You said I CAN assume all currents at all nodes are exiting but then you said that I have to heed the direction of the given i.

Maybe I'm phrasing my question in a strange way, let me try the math. Can I do this:

Node 1:

(60 - V1)/2 + (V2-V1)/8 + (0-V1)/10 = 0

Yes, fine.

Node 2:

(60 - V2)/10 +(V1-V2)/8 + (0-3i) = 0

Would this be correct?

All but the last term is fine. You're summing currents coming into V2, and the 3i is DEFINED to be coming into V2 because it is a current source and there's no choice when dealing with a current source. The other currents you can choose their directions arbitrarily because they are unknowns at this point. The current source, however, has a strictly defined direction on the diagram. So make the last term: 3i.

 

[Feodalherren]

Gotcha. Thank you very much!