# Understanding Nodal Analysis - Are nonreference nodes given an arbitrary voltage until found?

Discussion in 'Homework Help' started by Kurt_051, Sep 10, 2016.

1. ### Kurt_051 Thread Starter New Member

Nov 11, 2015
6
0
I'm relearning about nodal analysis when no power sources are present, and one of the things my book says is:

since resistance is a passive element,
by the passive sign convention, current
must always flow from a higher potential
to a lower potential.

It then proceeds to show an example of applying KCL and Ohm's Law to nonreference nodes, but my confusion is that when two nonreference nodes with a resistor in between had to be dealt with, one of the nonreference nodes was seemingly arbitrarily selected as the higher voltage

$i = \frac{v_{higher} - v_{lower}}{R}$

Without anything I can think of indicating that the node being subtracted was a lower voltage than the other. Are nonreference nodes arbitrarily selected as higher or lower in relation to other nonreference nodes, and that's ok as long as you are consistent? Is that what is going on here? E.g. if I found all the nonreference nodes in a circuit, would, for nodal analysis, I be able to put all of the node voltages in an arbitrary chain of inequalities like

$v_a < v_b < \ldots < v_i$

and this would be ok as long as all the equations between nodes were consistent with the chain?

2. ### DGElder Member

Apr 3, 2016
345
85
You don't need to know which voltage is greater, the assumption you make when setting up your KCL equations is the direction of the currents. Then those assumed current directions determine which voltage is assumed to be greater- using the PSC.

$i = \frac{v_{higher} - v_{lower}}{R}$

In this example, the "higher" voltage is determined by the assumed current direction through the resistor (via the PSC). The current direction flows from the higher voltage potential to the lower potential through the resistor.

To expand: If the resistor is between node 1 and node 2 and I chose to assume all the currents flow into node 1 when writing the node 1 equation, then the joining branch current would be expressed as (V2-V1)/R. If I had chosen all the currents as flowing out of node 1 then that current would be expressed as (V1-V2)/R.

Because each nodal equation is linearly independent, I can choose all the current directions to be into or out of node 2 without regard to what directions I chose for node 1. If I chose current directions for node 2 which had current moving in the opposite direction through the joining resistor that would be OK. Compared to a consistent current direction through R, that would be the equivalent of multiply the node 2 equation by -1. Since X+Y=0 and so does -1(X+Y)=0, the solutions for X and Y will be the same.

Now you have to more careful with assumed current directions in more complicated circuits with dependent sources, etc. and you will want to choose current directions which conform with given current directions, simplify analysis and avoid confusion, but I am just trying to drive home my point.

If this is not clear post a circuit example (with labeled elements and nodes for easy discussion) where you are having trouble with interpreting KCL.

Last edited: Sep 10, 2016
3. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490

Hi,

The short answer is that current is polarized as well as voltage, and the two are related to each other through the passive sign convention.

Thus if you have current flowing left to right and you call that positive, then the left hand voltage must be higher than the right hand voltage, so the left hand voltage is positive, the right hand voltage is negative, and the current is positive. Note that we had to consider THREE things there: two voltage polarities and one current polarity. It's not JUST about the two voltages, the current flow polarity must also be included in the reasoning.

Since most circuits are drawn with either horizontal or vertical elements, an exercise would be to write the equations for the current in a horizontal or vertical element where:
1. Current flows from left to right: (Vleft-Vright)/R=i
2. Current flows from right to left: (Vright-Vleft)/R=i
3. Current flows down: (Vtop-Vbottom)/R=i
4. Current flows up: (Vbottom-Vtop)/R=i

If you can write those four equations then you can write equations for almost any circuit. In circuits that are not drawn to that standard, if you orient the parts horizontally or vertically, then you can write for any circuit.
For example if you have a part drawn diagonally, you can orient that either horizontally or vertically or else just practice four more cases. Most circuits have them drawn horizontally or vertically though.

So practice those four cases and you'll see how easy it gets.
If you change the orientation of the current (make it negative) in any of those four cases, then you have to swap the two voltages. You dont normally have to do this, but just in case you do:
(Vleft-Vright)/R=i
(Vright-Vleft)/R=-i
so
(Vleft-Vright)/R=-(Vright-Vleft)R

Stick with the four cases first and get accustomed to that before moving on.

When you write your own circuit analysis software you have to learn the mechanics behind all of this stuff. I write my own software so i have to look at this stuff in detail.

Last edited: Sep 10, 2016
4. ### WBahn Moderator

Mar 31, 2012
17,748
4,796

The problem is in your nomenclature. You distinguish between the two voltages that make up the voltage difference across the resistor, but you don't distinguish that the current can flow in either of two directions through the resistor. It is not enough to say that the current in the resistor is 5 A, in order to be complete you need to indicate which direction that current is flowing. You do this by assigning a reference direction to the current -- THAT is what the passive sign convention is all about.

If a resistor is connected to nodes A and B, then the current flowing from A to B through the resistor is
$i_{AB} = \frac{v_A - v_B}{R}$
Similarly, the current flowing from B to A through the resistor is

$i_{BA} = \frac{v_B - v_A}{R}$

Either of these are equally correct and you can use either or both of them as you work a problem -- but the subscripts are critical since

$i_{BA} = - i_{AB}$

5. ### RBR1317 Active Member

Nov 13, 2010
232
48
It certainly seems that the theory of nodal analysis is much more complicated than the actual practice of writing nodal equations needs to be. My recommendation is to write each node equation as though that node's voltage is the highest voltage. So the first thing you write for each term in the node equation is that node's voltage. Forget about guessing in which direction branch currents flow because how can one know what is correct until the node equations are solved. If a node has current sources connected to it, then current flow away from the node is positive while current flow into the node is negative. This is the simplest possible method for writing node equations and the least error-prone.

6. ### WBahn Moderator

Mar 31, 2012
17,748
4,796
You don't have to guess which direction the currents are flowing or have to assume that each node's voltage is the highest voltage. You simply apply KCL at each node in the form of "the currents exiting the node sum to zero", although a more efficient way is to use the form that the sum of the currents exiting the node through passive elements equals the sum of currents entering the node through active sources. If the sources are current sources, then this this is trivial. If the active sources are voltage sources, then you need to be a bit more careful as this involves, formally, the notion of supernodes, though these can often be dealt with informally very easily.

7. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490
Hi,

There are actually a couple different ways of doing something like this, and a few more theories out there that we dont usually use, but i tend to be a little dogmatic myself here because i usually do the circuit the way i see it might work in the real world. For example, if i see a source on the left and a resistor on the right, i'll assume the current flow at the top of the source is out of the source and into the resistor. It just makes sense to use that idea even though we dont have to because it makes sense that the source is powering the resistor and using conventional current flow that means out of the positive terminal of the source and into the resistor. We have to do it some way or another so i like to try to keep it 'real', or as Larry Wilmore would have said before his show was canceled, "Keep it 100 percent"
On the other hand, in a computer program i do it as abstract as needed in order to make the code easier to write. By 'easier' i mean a little bit easier anyway