circuit with 2 voltage sources

PsySc0rpi0n

Joined Mar 4, 2014
1,755
But it isn't. The voltage across the right voltage source is +15 V while the voltage at the top right node (of your schematic with the components swapped) is -15 V due to the orientation of the supply.

I don't know what you mean by wanting to find another "mnemonic" to perform the analysis. If you mean finding better labels, that's a simple matter of just choosing different labels for things that are different (regardless of whether they might numerically have the same value). In this case, call the two voltage sources Vs1 and Vs2 and call the node voltages V1, V2, etc. (or Va, Vb, etc.).
The voltage source value is positive or negative depending on the way we measure it, I guess. If I place the black probe of a multmeter in the minus side and the red probe at the plus sign, it will read +15V, if I measure it the other way around, it will read -15V.

About the mnemonic for nodal analysis, I mean the mnemonic that I use to perform nodal analysis which is:

voltage at the required node minus voltage of branch 'n' over the impedance of branch 'n' equals the voltage at required node minus the voltage of branch 'n + 1' over the impedance of branch 'n + 1' and so on, having in mind the directions of currents I chose.

The source voltage is the voltage difference between the voltage at the labeled + terminal and the voltage at the labeled - terminal, i.e. V+ minus V-. The voltage of a node is the absolute voltage of a point in the circuit with respect to your chosen reference point, usually a common return or ground. As Bahn said, clearly and separately labeling sources and nodes will reduce confusion and the possibility of conflating the two.
I say the same... The value of the voltage of the voltage source depends on the way we measure it! At least is this way I think! Anyway, I got the point!
 

WBahn

Joined Mar 31, 2012
29,976
The voltage across a voltage source is defined as the voltage at the positive terminal minus the voltage at the negative terminal.

Sure, if you swap the leads of your voltmeter then you can reverse the polarity of ANY voltage reading. But when you say that the voltage of a source is X volts, that means that the voltage of the positive terminal minus the voltage of the negative terminal is X volts.

As for the mnemonic, it sounds like you are relying on memorizing a saying instead of understanding what nodal analysis IS and, from that, how to perform it.
 

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Yes, about the nodal analysis, there are a few considerations I have to memorize because things are not as straight forward as they look! I don't know any other of performing a nodal analysis without using my mnemonic!
 

WBahn

Joined Mar 31, 2012
29,976
Yes, about the nodal analysis, there are a few considerations I have to memorize because things are not as straight forward as they look! I don't know any other of performing a nodal analysis without using my mnemonic!
It's far better to understand what nodal analysis is -- nothing more than the systematic application of KCL. You are simply summing up the currents leaving a given node. So what's the current leaving node A to through resistance R to node B? It's (Va-Vb)/R. Do you see how that maps directly to you mnemonic? If you have this understanding, then you don't have to memorize some approach by wrote. You also don't have to worry about how to deal with any current sources that are connected to the node as it is obvious what to do with them.
 

PsySc0rpi0n

Joined Mar 4, 2014
1,755
It's far better to understand what nodal analysis is -- nothing more than the systematic application of KCL. You are simply summing up the currents leaving a given node. So what's the current leaving node A to through resistance R to node B? It's (Va-Vb)/R. Do you see how that maps directly to you mnemonic? If you have this understanding, then you don't have to memorize some approach by wrote. You also don't have to worry about how to deal with any current sources that are connected to the node as it is obvious what to do with them.
Sure, but the problem comes when we have voltage sources in opposite directions or when current directions makes the current equations signs to change! I understand that nodal analysis is just the sum of all currents flowing into and out a node. The problem is those situations when voltage sources orientations or current directions changes signs in the equations...
 
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WBahn

Joined Mar 31, 2012
29,976
It's not the sum of all current flowing THROUGH a node, it is the sum of all currents flowing OUT OF a node. The sign of a current flowing THROUGH a node is completely ambiguous. The sign of a current flowing OUT OF a node is not. If the current is flowing OUT of the node it is positive. If the current is flowing INTO a node it is negative. Period.
 

dannyf

Joined Sep 13, 2015
2,197
Sure, but the problem comes when we have voltage sources in opposite directions or when current directions makes the current equations signs to change! I
You are making a simple task overly complicated.

Two and only two principles of nodal analysis:

1. One node can only be at one potential, regardless of which branch you take to get there.

2. All current flowing in and out of a node should be balanced (summing up to zero).

Both are very intuitive.

So it doesn't matter the least bit if you assign negative sin to current into a node or out of a node, as long as they are consistent.

Anything beyond that isn't necessary.
 

PsySc0rpi0n

Joined Mar 4, 2014
1,755
I have edited my last post! Yes, I used some wrong words there. Sorry for that!

And I can agree with everything as said but the most important word said was "consistent"! This consistency goes away as soon as voltage sources turned up side down appear in a circuit! I know that I'm complicating, but as long as I don't understand it, I'll will always complicate it!

So, if I chose the original circuit form, the top right corner of the circuit cannot be a node, because from V3 to Vs2 negative terminal is only a wire making the top right corner of the circuit the same point as V3 or Vs2 negative terminal.
And in this case, isn't V3 voltage known? Isn't it imposed by Vs2? This is actually a question that I have if I don't swap Vs2 with the resistor in the same branch!

upload_2016-6-21_20-30-18.png
And if I perform nodal analysis "blindly" I would write something like:

(V3 - V2) / R = V3 / R + (V3 - Vs2) / R

But this would raise some questions to myself.
Vs2 is not a node, so, theoretically, this would be no nodal analysis anymore!
I wouldn't be sure about equation signs. I would ask to myself:

If I write:
(V2 - V3) / R = V3 / R + (V3 + Vs2) / R
or
(V2 - V3) / R = V3 / R + (Vs2 - V3) / R

Why would they be wrong? Or which one would be correct and why?
Even if I try to use the following reasoning, I'll get a consistent mnemonic to perform nodal analysis:
"current always flows from higher potentials to lower potentials (conventionally)"...
because I'll get to a point where I might not be able to judge where is higher and lower potentials!
 

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Jony130

Joined Feb 17, 2009
5,487

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PsySc0rpi0n

Joined Mar 4, 2014
1,755
Sorry @Jony130 ... V2 would be the former V2 node I assumed in the previous form of the circuit with Vs2 and R swapped!

So the 1st thing that is for sure is that the operand between the 2 node voltages considered in each term is minus...
Then, the voltage nodes values are the ones that might change. If the 2nd them of each equation term is referred to a voltage source that has the negative terminal pointing to the unknown node, then, that 2nd term will come with a minus like the one written as ...(V3 - (-Vs2) ) / R = 0. As Vs2 has it's negative terminal pointing to V3, then this value will be negative!

If this is correct I might adopt this new mnemonic!
 

WBahn

Joined Mar 31, 2012
29,976
I have edited my last post! Yes, I used some wrong words there. Sorry for that!

And I can agree with everything as said but the most important word said was "consistent"! This consistency goes away as soon as voltage sources turned up side down appear in a circuit! I know that I'm complicating, but as long as I don't understand it, I'll will always complicate it!

So, if I chose the original circuit form, the top right corner of the circuit cannot be a node, because from V3 to Vs2 negative terminal is only a wire making the top right corner of the circuit the same point as V3 or Vs2 negative terminal.
And in this case, isn't V3 voltage known? Isn't it imposed by Vs2? This is actually a question that I have if I don't swap Vs2 with the resistor in the same branch!

View attachment 108058
The only thing that is know from the schematic is that V3 is 15 V lower than the bottom-right node. But since we don't know the voltage of the bottom-right node (because we don't know the current in the bottom resistor) we don't know V3.

Remember, a voltage source merely establishes a voltage difference between the two nodes it is connected to.
 

WBahn

Joined Mar 31, 2012
29,976
But this would raise some questions to myself.
Vs2 is not a node, so, theoretically, this would be no nodal analysis anymore!
I wouldn't be sure about equation signs. I would ask to myself:
This is where understanding what nodal analysis IS instead of just trying to apply some memorized "mnemonic" comes into play.

The mnemonic you are trying to use is based on KCL stated simply as the sum of all currents leaving a node must sum to zero.

Consider the following:

Nodal1.png

For node V3, the currents leaving it must sum to zero:

I1 + I2 + I3 = 0

I1 = (V3 - V1)/R
I2 = (V3 - 0 V)/R = V3/R

For I3, notice that because the 15 V source and the bottom resistor are in series that we can write I3 for the bottom resistor knowing that it is equal to the I3 leaving node V3.

I3 =(V2 - 0 V)/R = V2/R

So we have

(V3 - V1)/R + V3/R + V2/R = 0

The voltage source imposes the relationship

(V2 - V3) = 15 V
V2 = V3 + 15 V

(V3 - V1)/R + V3/R + (V3 + 15 V)/R = 0

V3(1/R + 1/R + 1/R) = V1/R - 15 V/R

3·V3 = V1 - 15 V

We know that V1 = 10 V, so

V3 = -5/3 V = -1.67 V
 

dannyf

Joined Sep 13, 2015
2,197
So, if I chose the original circuit form,...
The world is much simpler than you think, if you simply remember why you are doing what you are doing, rather than what you are doing. there is no point remembering the ways of solving a particular problem. If you remember the principles used in solving that one particular problem, you can solve any problem, even if they are dissimilar.

the problem you mentioned can be solved easily if you simply follow the two principles laid out for you earlier.
 
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