HELP NEED SOME ADVICE

WBahn

Joined Mar 31, 2012
29,979
At node e:

node e: (6A) + (Vc-Vb)/(2 Ω) + (Ve-Vb)/(2 Ω) = 0

You counted the two terms that overlap though, so why wouldn't you do it for Node f?
Sorry I don't see it.
Because I made a mistake in transcribing from your expression and then correcting it in Post #12. I'm having to scroll the screen to see the diagram and then to see my work and my memory is getting bad enough that I shouldn't do that.

Node e: (6A) + (Vc-Vb)/(2 Ω) + (Ve-Vf)/(5 Ω) = 0

That should look much better. Do you agree with this one? That makes the set of equations:

Node a: Va = Vg + 40V
Node b: (Vb - Va)/(4 Ω) + (Vb - Vc)/(2 Ω) + (Vb - Vd)/(3 Ω) = 0
Node c: Vc = Ve + 40V
Node d: (Vd - Vb)/(3 Ω) - 6A + (Vd - Vh)/(1 Ω) = 0
Node e: (6A) + (Vc - Vb)/(2 Ω) + (Ve - Vf)/(5 Ω) = 0
Node f: (Vf - Vg)/(1 Ω) + (Vh - Vd)/(1 Ω) + (Vf - Ve)/(5 Ω) = 0
Node g: (Va - Vb)/(4 Ω) + (Vg - Vf)/(1 Ω) = 0
Node h: Vh = Vf + 30V

In checking this, what I did was first and foremost duplicated the page so that I can see the diagram and equations at the same time. Then I went through node by node and put my finger on the node and verified the term for each branch starting with the branch leaving the node going to the left and proceeding clockwise, reordering the terms if necessary. Of course, this doesn't apply to the nodes governed by an auxiliary equation. The mistake you flagged is the only once I found. Next I defined Node g as the reference node again and substituted in all of the auxiliary equations to get:

Node b: (Vb - ((0V) + 40V))/(4 Ω) + (Vb - (Ve + 40V))/(2 Ω) + (Vb - Vd)/(3 Ω) = 0
Node d: (Vd - Vb)/(3 Ω) - 6A + (Vd - (Vf + 30V))/(1 Ω) = 0
Node e: (6A) + ((Ve + 40V) - Vb)/(2 Ω) + (Ve - Vf)/(5 Ω) = 0
Node f: (Vf - (0V))/(1 Ω) + ((Vf + 30V) - Vd)/(1 Ω) + (Vf - Ve)/(5 Ω) = 0

The next step is to expand these out and collect terms. I'm going to do this in four steps to minimize the risk of introducing bookkeeping errors. First I am going to get rid of inner parens in the numerators as well as any 0V terms. Second I'm going to do is multiply each equation by whatever resistance will clear out the denominators or, in other words, the least common multiples of the denominator resistances. Third I'm going to distribute the multipliers in each term, and finally I'm going to collect terms in a consistent order, introducing null terms as needed so that each equation has all four unknown terms in it. Not only does this reduce the chance of making a mistake, but it makes it a lot easier to track down and correct a mistake if I do make one. Pay particular attention to how the units are treated in each step.

Step #1:

Node b: (Vb - 40V)/(4 Ω) + (Vb - Ve - 40V)/(2 Ω) + (Vb - Vd)/(3 Ω) = 0
Node d: (Vd - Vb)/(3 Ω) - 6A + (Vd - Vf - 30V)/(1 Ω) = 0
Node e: (6A) + (Ve + 40V - Vb)/(2 Ω) + (Ve - Vf)/(5 Ω) = 0
Node f: (Vf)/(1 Ω) + (Vf + 30V - Vd)/(1 Ω) + (Vf - Ve)/(5 Ω) = 0

Step #2:

Node b: (Vb - 40V)(3) + (Vb - Ve - 40V)(6) + (Vb - Vd)(4) = 0 (multiplied by 12 Ω)
Node d: (Vd - Vb)(1) - 6A(3 Ω) + (Vd - Vf - 30V)(3) = 0 (multiplied by 3 Ω)
Node e: (6A)(10 Ω) + (Ve + 40V - Vb)(5) + (Ve - Vf)(2) = 0 (multiplied by 10 Ω)
Node f: (Vf)(5) + (Vf + 30V - Vd)(5) + (Vf - Ve)(1) = 0 (multiplied by 5 Ω)

Step #3:

Node b: 3Vb - 120V + 6Vb - 6Ve - 240V + 4Vb - 4Vd = 0
Node d: Vd - Vb - 18V + 3Vd - 3Vf - 90V = 0
Node e: 60V + 5Ve + 200V - 5Vb + 2Ve - 2Vf = 0
Node f: 5Vf + 5Vf + 150V - 5Vd + Vf - Ve = 0

Step #4:

Node b: + 13Vb - 4Vd - 6Ve - 0Vf = 360 V
Node d: - 1Vb + 4Vd - 0Ve - 3Vf = 108 V
Node e: - 5Vb - 0Vd + 7Ve - 2Vf = -260 V
Node f: - 0Vb - 5Vd - 1Ve + 11Vf = - 150 V

It's worth pointing out that in my Node f equation above I originally had 10Vf. But I am in the habit of working out the equations and then going back and reviewing them term by term to see if I got everything and, in that case, I had missed the next to last term. Checking your work as you go will save you SO much time in the long run -- get in the habit of doing it.

At this point there is a very useful check that can be done. Notice that in each equation the term for that node's voltage is positive while all of the other voltage terms are negative. That's a result of developing each equation based on the current flowing away from that node.

Now you can use whatever technique you like to solve this system of linear equations. Since I seldom have more than three, I usually just do it by hand. But I decided to take this opportunity to teach myself how to use Excel's matrix functions to do it and got the following:

Vb = 32.87 V
Vd = 36.55 V
Ve = -13.157 V
Vf = 1.7831 V

Are these correct? I don't know. The point is that I'm not just going to accept whatever is spat out at me at face value, but instead I'm going to check it. What if you hadn't pointed out the error I made? I would probably have proceeded to work with those equations and gotten a set of results that looked just as reasonable as these. That is why you always check your results. The beauty of most engineering problems is that the validity of the results can be confirmed from the results themselves. First, let's get the rest of the node voltages using the auxiliary equations:

Va = 40V
Vb = 32.87 V
Vc = 26.84 V
Vd = 36.55 V
Ve = -13.157 V
Vf = 1.7831 V
Vg = 0V
Vh = 31.78 V

The quickest check is probably to see if the currents in the lefthand loop are consistent.

(Va - Vb)/(4 Ω) = 1.7825 A
(Vf - Vg)/(1 Ω) = 1.7831 A

Close enough and a very good sign. But not done yet. Next let's get the currents flowing into Node d from Node b and f and see if those add to -6A to be consistent with the current coming from Node e.

(Vb-Vd)/(3 Ω) = -1.2267 A
(Vh - Vd)/(1 Ω) = -4.770 A
Sum = -5.997 A

Finally, we'll do the same for the currents coming into Node e.

(Vb-Vc)/(2 Ω) = 3.015 A
(Vf - Ve)/(5 Ω) = 2.988 A
Sum = 6.003 A

Since we believe in being thorough, and because the info we need is in the work above, let's also verify that KCL is actually satisfied at Nodes b and f, too.

Node b: 1.7825 A ?= -1.2267 A + 3.015 A = 1.7883 A (yes)
Node f: 1.7831 A + -4.770 A + 2.988 A = 0.0011 A?= 0 (yes)

So these are the correct answers for the node voltages. Notice that in checking our work, we actually calculated the I1 and I2 currents that were originally asked for.
 

WBahn

Joined Mar 31, 2012
29,979
I think I see what you did now, you used super position
No, I just made a mistake that you caught because you were trying to make sense of what I wrote. Good catch. That is why I went ahead and worked the rest of the problem for you. Look that over and get comfortable with it, then I will show you how to get the necessary equations MUCH quicker.
 

upopads2

Joined Sep 20, 2014
54
No, I just made a mistake that you caught because you were trying to make sense of what I wrote. Good catch. That is why I went ahead and worked the rest of the problem for you. Look that over and get comfortable with it, then I will show you how to get the necessary equations MUCH quicker.
Sorry for not doing very much....I should be able to do this stuff easier than I do.
 

WBahn

Joined Mar 31, 2012
29,979
Sorry for not doing very much....I should be able to do this stuff easier than I do.
Nothing to be sorry for. I made the mistake and that resulted in you chasing a rabbit down a hole for a bit. This stuff isn't easy at first and it takes a lot of practice working problems to get it down.
 

WBahn

Joined Mar 31, 2012
29,979
Current coming from node e into node d is 6A not -6A but it does not matter, the value is the same.
I agree that the current coming from Node e into Node d is 6A and not -6A and it does matter because if you settle for getting the magnitude right but ignore a sign mismatch, then that is a VERY wrong answer. So when you see something like that it is important to resolve it.

What I wrote is correct, but I see where you are getting confused, so let's look at it (assuming I've identified the right part of my post that you are referring to)

WBahn said:
Close enough and a very good sign. But not done yet. Next let's get the currents flowing into Node d from Node b and f and see if those add to -6A to be consistent with the current coming from Node e.
(Vb-Vd)/(3 Ω) = -1.2267 A
(Vh - Vd)/(1 Ω) = -4.770 A
Sum = -5.997 A
The current flowing into Node d from Node b and Node f have to add up to -6A. This is because, when added to the +6A flowing from Node e to Node b the result has to be zero. The key is that I didn't say that they had to add to -6A to be EQUAL to the current coming from Node e (which is +6A), but that they had to add to -6A in order to be CONSISTENT with the current coming from Node e.
 
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WBahn

Joined Mar 31, 2012
29,979
Hopefully you have taken the time to become comfortable with all of the steps in the analysis above, because understanding them will be key to understanding how they are massaged to do the following:

So the way that we analyzed the circuit last time is very close to just applying KCL directly, which is fine and it works, but you saw the amount of work that was involved to deal with the bookkeeping. So the question is whether we can formalize this approach a bit and come up with a means of getting right to the heart of the matter and, as much as possible, writing the equations down by inspection.

The answer is, "Yes," and the formalism is known as NVA - "Node Voltage Analysis", or often just called "Nodal Analysis". It is really nothing more that a consist means of applying KCL at the essential (what I called "critical" before) nodes.

If we only have resistors connecting a Node N to the adjacent nodes:

For Node N, add Vn multiplied by the conductances connecting it to other nodes (remember that conductance is 1/resistance). Then subtract the voltages at the adjacent nodes multiplied by the conductance connecting them to Node N. We write these in the order of the nodes. Since this represents all of the current leaving the node and there are no other paths, we set this equal to zero.

In our circuit, an example of this would be Node b.

- Va[(1/4Ω)] + Vb[(1/4Ω)+(1/2Ω)+(1/3Ω)] - Vc[(1/2Ω)] - Vd[(1/3Ω)]= 0

It's important that you realize that this is just a reordering of the KCL expression for this node:

(Vb-Va)[(1/4Ω)] + (Vb-Vc)[(1/2Ω)] + (Vb-Vd)[(1/3Ω)] = 0

Now look at how you can sanity check the equation. Vb is positive while all of the other terms are negative. Furthermore, each conductance that appears in the Vb term appears exactly once in the other terms. If you get in the habit or writing them in the same order in the Vb term as the other terms appear, it makes performing the sanity check very easy.

The next simplest case is if we have a current source coming into a node. In this case we first deal with the resistances connected to the node exactly as we did before, but now we have on the left hand side the currents going out through the resistors and we set this equal to the current coming in through the current sources on the right hand side.
In our circuit, an example of this would be Node d.

- Vb[(1/3Ω)] + Vd[(1/3Ω)+(1/1Ω)] - Vh[(1/1Ω)] = 6A

Next we have the case where we have a voltage source connected to a node. This is the tricky case and is often a reason to choose another analysis method, although dealing with it using NVA is not as bad as it seems at first. As we have seen, the trick is to use the node on the other side of the voltage source in order to get at the currents. This approach is known as using a "supernode" because what we are effectively doing is writing the equation for a small collection of nodes instead of for a single node.

In our circuit, an example of this would be Node f.

- Vd[(1/1Ω)] - Ve[(1/5Ω)] + Vf[(1/1Ω)+(1/5Ω)] - Vg[(1/1Ω)] + Vh[(1/1Ω)] = 0

Notice here that the nodes that make up our supernode are Node f and Node h and those two node voltages are positive while the others are negative. Also note the each conductance appears exactly twice, once in the term for an external node (a negative term) and once in the term for an internal node (a positive term).

These are the basic three cases, but in the examples we have ignored the distinction between essential and non-essential nodes. That's fine, but it means that we will need to include all of the auxiliary equations that go along iwth the non-essential nodes and then substitute them in.

However, with practice, we can actually do this by inspection. The key is to first identify the essential nodes, which are the nodes that connect more than two branches. In our case these are Nodes {b,d,e,f}. Then pick one of these nodes to be the reference node (let's pick Node f). Now we want to write the node equations for the remaining essential nodes and use the notion of a supernode to include any adjacent non-essential nodes and to write the node voltages for any non-essential nodes in terms of the essential nodes and attached voltage sources. This is a lot to absorb, but with practice it will start to make a lot of sense.

So with all of this in mind, here is where I would start out when working this problem:

START OF ANALYSIS

Code:
Node b: + Vb[(1/5Ω)+(1/3Ω)+(1/2Ω)] - (0V+40V)[(1/5Ω)] - Vd[(1/3Ω)] - (Ve+40V)[(1/2Ω)] = 0
Node d: - Vb[(1/3Ω)] + Vd[(1/3Ω)+(1/1Ω)] - (0V+30V)[(1/1Ω)] = 6A
Node e: - (Vb-40V)[(1/2Ω)] + Ve[(1/2Ω)+(1/5Ω)] - 0V[(1/5Ω)] = -6A
It's entirely possible to directly write the terms involving the voltage sources on the right side of the equation, but I find that it is easier for me to keep things straight by putting them on the left side first and then moving them over at the same time that I clear out the denominators by multiplying the entire equation by an appropriate resistance. If the equations are much messier than these here, I will do these things in separate steps. I also tend to include the 0V (i.e., Vf) terms explicitly as it helps be with the sanity checks and then remove those in the next step, as well.

Code:
Node b: + 31Vb - 10Vd - 15Ve = 240V + 600V =  840V
Node d: -   Vb +  4Vd        =  18V + 90V  =  108V
Node e: -  5Vb        +  7Ve = -60V - 200V = -260V
Solving this we get:

Vb = 31.08 V
Vd = 34.77 V
Ve = -14.940 V

END OF ANALYSIS

Now, if you compare these to our prior results, you might think that there is a mismatch. Before we got:

Va = 40V
Vb = 32.87 V
Vc = 26.84 V
Vd = 36.55 V
Ve = -13.157 V
Vf = 1.7831 V
Vg = 0V
Vh = 31.78 V

But remember that before we defined Vg to be 0V and this time we defined Vf to be 0V. This results in a voltage shift by Vf, meaning that in this new analysis all of the voltages should be 1.7831V lower than they were in the old analysis. If we add 1.7831V to the new values, we get:

Vb = 31.08 V + 1.7831 V = 32.86 V
Vd = 34.77 V + 1.7831 V = 36.55 V
Ve = -14.940 V + 1.7831 V = -13.157 V

You can't ask for much better agreement than that.

Hopefully you will agree that taking the time to get competent at Nodal Analysis will provide you with an extremely powerful analysis tool.

Next we can look at using KVL and Loop Analysis and then formalizing that to MCA - Mesh Current Analysis.
 

upopads2

Joined Sep 20, 2014
54
I'm impressed how you got Bill to do so much work for you! It certainly shows he is more generous / helpful than you originally thought.
Yup and I certainly feel like an *** for thinking that initially and believe me. Though you only can think about what you see. If nothing else from studying his problem here what hits home the most is when he says "always always, always". I'll never forget that. Never ever...
 
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