#### John Ramelb

Joined Feb 2, 2014
26
give me some clue how i can manipulate this cu rrent source back to dc source has anyone known? thanks

#### Jony130

Joined Feb 17, 2009
5,211
Try to use a source transformation and convert voltage source with series resistors into a current source with parallel resistors.

#### WBahn

Joined Mar 31, 2012
26,398
It IS a DC source. Are you trying to transform it into an equivalent voltage source?

Joined Sep 20, 2014
54
I'm going to completely embarrass myself and attempt to write loop equations for this....Which WILL BE COMPLETELY WRONG but at least you have something to show you tried.....

I1 = 40V/(4Ohms) + (40V)/(3Ohms) -I2 +6A + (40V/1Ohm)+ (40V-30V/1Ohm) -I4 -I3
I2= -I1 -6A (-30V/1Ohm) -(30V/3Ohm)-(30V/2Ohm)-[(30V-40V)/5Ohms]
I3=-I1 +I2 +6A -(40V/5Ohms)
I4=30V/6Ohm -6A -I3

#### shteii01

Joined Feb 19, 2010
4,644
I'm going to completely embarrass myself and attempt to write loop equations for this....Which WILL BE COMPLETELY WRONG but at least you have something to show you tried.....

I1 = 40V/(4Ohms) + (40V)/(3Ohms) -I2 +6A + (40V/1Ohm)+ (40V-30V/1Ohm) -I4 -I3
I2= -I1 -6A (-30V/1Ohm) -(30V/3Ohm)-(30V/2Ohm)-[(30V-40V)/5Ohms]
I3=-I1 +I2 +6A -(40V/5Ohms)
I4=30V/6Ohm -6A -I3
The OP is long gone.

Joined Sep 20, 2014
54

#### WBahn

Joined Mar 31, 2012
26,398
I'm going to completely embarrass myself and attempt to write loop equations for this....Which WILL BE COMPLETELY WRONG but at least you have something to show you tried.....

I1 = 40V/(4Ohms) + (40V)/(3Ohms) -I2 +6A + (40V/1Ohm)+ (40V-30V/1Ohm) -I4 -I3
I2= -I1 -6A (-30V/1Ohm) -(30V/3Ohm)-(30V/2Ohm)-[(30V-40V)/5Ohms]
I3=-I1 +I2 +6A -(40V/5Ohms)
I4=30V/6Ohm -6A -I3
So, YOU are going to supply loop equations, right or wrong is irrelevant, so that the OP will have something to show that THEY tried?

Leaving aside the ethics involved, having loop equations doesn't address the question the OP appears to have been asking. Second, giving equations that have two undefined currents (I3 and I4) doesn't help much. Third, equations that are dimensionally inconsistent are guaranteed to be wrong: (40V - 30V/1Ohm) is volts - amps, and 6A (-30V/1Ohm) is amps-squared. I suspect that both of these are just sloppy mistakes, but you need to check your work better (and, yes, we all make sloppy mistakes and, no, we don't always catch them, but we can catch the overwhelming majority of them by getting in the routine habit of sanity checking ALL of our work). I do want to give you kudos, though, for showing your units properly.

Okay, since the OP is almost certainly long gone (or at least the homework is long past due) I'll accept your hijack of their thread as now being YOUR efforts to solve the problem for YOUR benefit and proceed accordingly.

First, don't worry about embarrassing yourself -- making efforts that don't pan out and then struggling with them is perhaps the best way to learn just about anything.

You have three meshes, and therefore only need three loop equations. There are seven possible loop equations, any three of which are independent with the remaining four being linear combinations of the other three. So, even if your equations are correct, one of them is redundant.

The bigger problem is that loop equations are fundamentally sums of voltages (they are just KVL), and your equations are sums of currents. Are you sure you aren't trying to write node equations? That seems likely since there are four non-trivial nodes in the circuit. Now, you still only have three independent node equations because you get to pick one of the nodes and define it to be your reference node (i.e., "ground", "common", or "0v").

About the only thing else I can offer until you revise your work (assuming you are interested in doing so), is to note that you are completely missing the boat on what Ohm's Law means. Ohm's Law relates the voltage across a resistance and the current through THAT resistance to the resistance of THAT resistance. So when you have 40V/(4Ohms), you are saying that 40V appears across a 4Ohm resistor. But look at the diagram. Neither 40V source appears across a 4Ω resistor. The same for nearly all of your other terms. You can't just pick whatever voltage you want and divide it by whatever resistance you want and get a current that has any meaning you want.

So let's set up one of the node equations properly as an example and then you can set up the others.

Let's first identify the non-trivial nodes by examining all eight. A node is non-trivial if it connects more than two components. If it only connects two, then the current coming in from one is equal to the current going out to the other and there is no useful information gained about the circuit as a whole. But if there are more than two, then the current can split and how it splits does provide useful information about the circuit as a whole.

Node a: Trivial
Node b: Non-trivial
Node c: Trivial
Node d: Non-trivial
Node e: Non-trivial
Node f: Non-trivial
Node g: Trivial
Node h: Trivial (this node is unmarked -- it is the junction of the 30 V source and the 1 Ω resistor)

So we need to write node equations for (i.e., apply KCL at) three of the non-trivial nodes. Let's pick Node b. Let's sum the currents going out of the node and set them equal to zero.

Node b: (Vb - Va)/(4 Ω) + (Vb - Vc)/(2 Ω) + (Vb - Vd)/(3 Ω) = 0

After writing the node equations for three of the non-trivial nodes, you will almost certainly have more than three unknown voltages. For instance, in just the equation above we have Va, Vb, Vc, and Vd. That's fine, but it means we will need additional constraint equations, also known as auxiliary equations, to deal with them. These are usually easy to come up with. For instance, we know that

Va = Vg + 40V
Vh = Vf + 30V
Vc = Ve + 40V

At this point we know that we have eight unknown node voltages and so we will need eight independent equations to solve for them. We know we get to arbitrarily define one of them to be 0V. We know that we will have three non-trivial node equations to work with, and we know that we already have three auxiliary equations. We only need one more. That one is more subtle, but it comes from the fact that we know that whatever current is flowing in the 4 Ω resistor from Node a to Node b is the same current that is flowing in the 1 Ω resistor from Node f to Node g since they are in series. Thus:

(Vb - Va)/(4 Ω) = (Vf - Vg)/(1 Ω)

Note that is is critical to get the directions correct.

Now you try to write the node equations for Nodes d, e, and f (go ahead and do all three for practice and we'll see how one is redundant).

The ball is in your court.

Joined Sep 20, 2014
54
node d: (Vd-Vb)/(3Ohms) -6A + (Vd-Vf)/(1 Ohm) = 0
node e: 6A+ (Ve-Vh)/(5Ohms) +(Ve-Vb)/(2Ohms) = 0
node f: (Vf-Ve)/(5 Ohms) +(Vf-Vg)/(1Ohms) + (Vf-Vd)/(1Ohm) = 0

How on earth do you know OP is gone?

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#### WBahn

Joined Mar 31, 2012
26,398
....I'm astonished by how much you posted I don't even really know why you're giving me help since i don't see you do it in other topics. But here you go, here's my unethical attempt to solve the problem:

node d: (Vd-Vb)/(3Ohms) -6A + (Vd-Vf)/(1 Ohm) = 0
node e: 6A+ (Ve-Vh)/(5Ohms) = 0
node f: (Vf-Ve)/(5 Ohms) +(Vf-Vg)/(1Ohms) + (Vf-Vd)/(1Ohm) = 0
I'm giving you help because you posted an effort to work the problem and I'm willing to help anyone that is willing to try.

The "ethics" part of the discussion is moot at this point and had only to do with blindly offering something to someone else in order for them to be able to submit YOUR work and claim it as their own for credit.

You are close, but not quite there:

node d: (Vd-Vb)/(3Ohms) -6A + (Vd-Vf)/(1 Ohm) = 0

The first two terms are correct, but the last term is not. (Vd-Vf) is NOT the voltage that appears across the 1 Ω resistor. Look a bit more carefully and try again.

node e: 6A+ (Ve-Vh)/(5Ohms) = 0

The first term is correct. The second term has a similar problem to the prior node equation. Are you sure you have Nodes f and h correct in your mind? Node f is marked on the diagram. Node h is the node between Node d and Node f along the center branch. You are missing a third term. Remember, each term in the equation represents the current flowing outward from the node along one of the available branches. For Node e, you have three available branches and therefore you need three terms. The fact that one of the branches is directly connected to a voltage source doesn't mean current can't flow that way and, therefore, you can't just ignore it. What you need to do is look at the other side of the voltage source and see if you can write the current flowing outward from Node e through the voltage source in terms of other branch components. In this case, consider the voltages at Nodes b and c and the 2 Ω resistor between them.

node f: (Vf-Ve)/(5 Ohms) +(Vf-Vg)/(1Ohms) + (Vf-Vd)/(1Ohm) = 0

The last term has the same problem identified in similar terms in the other two. So you're actually very close. I think you'll probably get it on the next attempt.

Oh, and here is a bit of a nit: You can insert the capital Omega symbol, Ω, using Alt-234 (by holding down the Alt key and using the number pad (NOT the number keys above the letter keys) to type 234). You can also copy paste it from elsewhere on the page. If you are going to type out "ohm" as the unit, then it should be lower case (that is a general rule for all units though there are a handful of exceptions) and there should be a space between the numerical value and the unit. Otherwise, 5Ohm is too easily mistaken for 50 ohms upon a casual glance. In general there is supposed to be a space between numbers and units, even if symbols are used, so it should be 5 Ω and not 5Ω. Admittedly, sometimes that makes things look a bit awkward and it can even lead to awkward line breaks, so sometimes it is reasonable to omit the space.

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Joined Sep 20, 2014
54
node d: (Vd-Vb)/(3Ohms) -6A + (Vd-Vh)/(1Ohm) = 0
node e: 6A - 40V +(Ve-Vb)/(2Ohms) = 0
node f: (Vf-Vg)/(1Ohms) + (Vf-Vd)/(1Ohm) - 30V = 0

I think this is it.

Joined Sep 20, 2014
54
Alt 234 does this to me ê ë ìí ï

#### WBahn

Joined Mar 31, 2012
26,398
node d: (Vd-Vb)/(3Ohms) -6A + Vd-Vh/1Ohm = 0
node e: 6A - 40V +(Ve-Vb)/(2Ohms) = 0
node f: (Vf-Vg)/(1Ohms) + (Vf-Vd)/(1Ohm) - 30V = 0

I think this is it.
Nope, but we'll get there.

node d: (Vd-Vb)/(3Ohms) -6A + Vd-Vh/1Ohm = 0

You have a typo here. Remember that multiplication and division take precedence over addition and subtraction, so what you have actually written is:

node d: (Vd-Vb)/(3 Ohms) - 6A + Vd - (Vh/1 Ohm) = 0

When what you meant to write is:

node d: (Vd-Vb)/(3 Ohms) - 6A + (Vd-Vh)/1 Ohm = 0

This is important for a few reasons. First, people have to work with what you write, not what you meant to write. Second, human are pretty good at inferring the correct things from context, even if it is technically incorrect as presented. This means that it is very possible that people will see what you wrote and interpret it the way you intended. Some of the time. But computers are very literal and if you were to type this into a program, a simulator, a spreadsheet, or you name it then it will interpret it exactly as written and not how you meant it to be written. So it becomes important to develop the habit of seeing expressions as they are, and not how you would like them to be, consistently. Finally, you might very well know what you meant in the line that you wrote, but there is a very good chance that as you proceed through your work you will lose the visual cues you (and possibly others) were relying on to "get it right" from the context and, at some point, start working with what is actually there, which is wrong. If you are tracking your units, then there is a good chance that you would catch the problem soon thereafter.

Anyway, adjusted for what I'm sure you meant to write, this is correct.

node e: 6A - 40V +(Ve-Vb)/(2Ohms) = 0

This one we KNOW is wrong because all we need to do is look at the units. Your first term is amps and your second term is volts. They CANNOT be added together.

For your second term, you know that you want the current flowing upward in the 40 V source. But notice that this is the same current that is flowing to the left through the 2 Ω resistor; that current is (Vc-Vb)/(2 Ω). So our equation should be:

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

Do you see how I came up with that?

node f: (Vf-Vg)/(1Ohms) + (Vf-Vd)/(1Ohm) - 30V = 0

Same thing here. The first and second terms are amps but the final term is volts. No can do. Your first term is correct, but your second term is claiming that (Vf-Vd) is the voltage across the 1 Ω resistor. It isn't.

Always, always, ALWAYS check your units. I routinely check my units every line or at least every two to three lines throughout my work. If I make a mistake, it will usually mess up the units. If I'm checking every line or two, then I seldom have to go back more than a line or two to find and fix most of the errors I make.

See if you can apply what I did to the Node e equation to fix your Node f equation.

#### WBahn

Joined Mar 31, 2012
26,398
Alt 234 does this to me ê ë ìí ï
I have no idea why it would give you four characters. What kind of keyboard are you using? You need one that has a separate numeric keypad. You can't use the keys above the letters nor can you use the numlock on laptop keyboards that lack a keypad and that turn a section of the letter keys into a numeric keypad (which really sucks). If you are using a smartphone or something like that, I have no idea but it would not surprise me that it doesn't work. So just keep typing out the units, but please do use lower case and do put a space between the numeric part and the units part.

Joined Sep 20, 2014
54
Why do you need both Ve-Vb and Vc-Vb?

node f: (Vf-Vg)/(1Ohms) + (Vf-Vh)/(1Ohm) + (Vh-Vd)/(1Ohm)+(Vf-Ve)/(5Ohms) = 0
or

node f: (Vf-Vg)/(1Ohms) + (Vf-30V)/(1Ohm) + (30V-Vd)/(1Ohm) + (Vf-Ve)/(5Ohms)= 0

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Joined Sep 20, 2014
54
I have no idea why it would give you four characters. What kind of keyboard are you using? You need one that has a separate numeric keypad. You can't use the keys above the letters nor can you use the numlock on laptop keyboards that lack a keypad and that turn a section of the letter keys into a numeric keypad (which really sucks). If you are using a smartphone or something like that, I have no idea but it would not surprise me that it doesn't work. So just keep typing out the units, but please do use lower case and do put a space between the numeric part and the units part.

ê

#### WBahn

Joined Mar 31, 2012
26,398
Why do you need both Ve-Vb and Vc-Vb?

node f: (Vf-Vg)/(1Ohms) + (Vf-Vh)/(1Ohm) + (Vh-Vd)/(1Ohm)+(Vf-Ve)/(5Ohms) = 0
or

node f: (Vf-Vg)/(1Ohms) + (Vf-30V)/(1Ohm) + (30V-Vd)/(1Ohm) + (Vf-Ve)/(5Ohms)= 0
You can't just say that Vh is 30V. That is only true if you have defined Vf to be 0V, and we haven't defined a reference at all. We could pick any node to be 0V and while Node f isn't unreasonable, arguably a better choice would be Node g.

Let's not lose track of what our equation means:

Node f: [The current flowing to the left from Node f toward Node g] + [The current flowing up from Node f toward Node h] + [The current flowing to the right from Node f toward Node e] = 0

You have THREE currents and therefore should end up with THREE terms, not four. You middle two terms are the same current, which means that you are double counting it. Think of it this way: You want to count all of the people leaving a building through three exits. At two of the exits you can count them directly, but at the third exit, for whatever reason, you can't. However, you know that every person that leaves the building has to enter an covered walkway and that only people that leave the building can enter that walkway. So instead of counting the people leaving the third exit, you count the people leaving the covered walkway. This will work, but you cannot count BOTH the people leaving the building by the third exit AND the people leaving the covered walkway, otherwise you are counting those people twice.

[The current flowing to the left from Node f toward Node g] = (Vf - Vg)/(1 Ω)
[The current flowing up from Node f toward Node h] = [The current flowing up from Node h toward Node d] = (Vh - Vd)/(1 Ω)
[The current flowing to the right from Node f toward Node e] = (Vf - Ve)/(5 Ω)

Hence, we have:

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

So let's write all of our node equations in one place:

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 - Vb)/(2 Ω) = 0
Node f: (Vf - Vg)/(1 Ω) + (Vh - Vd)/(1 Ω) + (Vf - Ve)/(5 Ω) = 0
Node g: (Vg - Vf)/(1 Ω) + (Va - Vb)/(4 Ω) = 0
Node h: Vh = Vf + 30V

Notice that we haven't written a node equation for Node g until now, so be sure that you are comfortable with how it was developed.

It now would appear that we have eight equations and eight unknowns, but appearances are deceiving. Only seven of these equations are independent, meaning that we can pick any seven of them and manipulate them to get the eighth. That certainly isn't obvious, but there are formal ways to test for independence and that is what they would show. So we really have seven equations and eight unknowns, but this is where our ability to arbitrarily choose one node to be at whatever voltage we want comes into play. We merely pick one of the node equations above and delete it (since we know it is redundant, we haven't lost any information) and instead declare the voltage at that node to be Vref. Vref can be anything. It can be -458.231V. But in almost all cases it makes the most since to declare it to be 0V. Since I didn't have you derive the equation for Node g, let's pick that one to get rid of and simply declare it to be 0V:

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 - Vb)/(2 Ω) = 0
Node f: (Vf - Vg)/(1 Ω) + (Vh - Vd)/(1 Ω) + (Vf - Ve)/(5 Ω) = 0
Node g: Vg = 0 V
Node h: Vh = Vf + 30V

Now we have eight independent equations in eight unknowns and we can use any method we want to solve them. But we know that we have four trivial nodes, namely {a,c,g,h}, so let's see if we can't get rid of those variables by substituting the equations for them into the other equations:

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 - Vb)/(2 Ω) = 0
Node f: (Vf - (0V))/(1 Ω) + ((Vf + 30V) - Vd)/(1 Ω) + (Vf - Ve)/(5 Ω) = 0

Now we have four equations in four unknowns. Had we chosen one of the non-trivial nodes to be 0V, we would have reduced that to three equations in three unknowns -- something to keep in mind for the future, huh?

Next we would expand these out and simplify them into a set of linear equations in our four variables and proceed to solve them, which I will leave up to you. If you give it a go, I will be more than happy to look your work over and will also give you some pointers, after the fact, on how you could make the equations easier to work with. I can then show you how you could write down the equations in one go by inspection, which is the real power of Nodal Analysis. After that, we could turn our attention to Mesh Analysis, if you are game.

Joined Sep 20, 2014
54
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.

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Joined Sep 20, 2014
54
When I hold down alt and press 2 3 4 i do the 2 3 4 synchronously and when i do that i get this: ê

#### WBahn

Joined Mar 31, 2012
26,398
When I hold down alt and press 2 3 4 i do the 2 3 4 synchronously and when i do that i get this: ê
What kind of computer and keyboard are you using?