# Thread: Kirchhoff's rules; Junction with 2 loops, or 3 loops

1. Hi everyone,

I've been having an on going debate with my physics professor about Kirchhoff's rules using the junction rule and loop rule.

I've been working on this off and on for a couple of days and can't really figure out why the circuit in the image below isn't giving the current flowing through each resistor using 3 loops. Loop1 in the circuit is the system as a whole, moving around the perimeter of the circuit, loop2 I used as being the left loop, and loop3 is the right loop.

In the image, at the very top, I get the correct answers using the loops 2 and 3 in conjunction with the Junction for point B, I1=.3, I2=.5, and I3=.2 respectfully.

I verified these answers with a problem out of a book.

Also I solved the system with 3 equations and three unknowns using a 3 X 4 Matrix.

But using 3 loops the answers are way different. I1=-.25, I2=.166, I3=0.

Anyways my professor swears that you can use the 3 loops to solve the currents.

Me personally I don't think its possible. It looks promising and should work I guess in theory, but I've tried it in every way I can think of and can't get it to work.

Can anyone tell me why you can't use 3 loops? There must be a reason. I was think it probably had something to do with conservation of forces on the outside loop1, but i'm unsure.

Thanks for the help,
rcowboy

Okay I'm having trouble getting the image to post but heres the URL. I'll keep working on it.

http://i673.photobucket.com/albums/v...y/scan0003.jpg

Well, can't get the image to post on forum, maybe someone can help with image. Anyways link is above.

2.

3. Originally Posted by rcowboy
Hi everyone,

I've been having an on going debate with my physics professor about Kirchhoff's rules using the junction rule and loop rule.

I've been working on this off and on for a couple of days and can't really figure out why the circuit in the image below isn't giving the current flowing through each resistor using 3 loops. Loop1 in the circuit is the system as a whole, moving around the perimeter of the circuit, loop2 I used as being the left loop, and loop3 is the right loop.

In the image, at the very top, I get the correct answers using the loops 2 and 3 in conjunction with the Junction for point B, I1=.3, I2=.5, and I3=.2 respectfully.

I verified these answers with a problem out of a book.

Also I solved the system with 3 equations and three unknowns using a 3 X 4 Matrix.

But using 3 loops the answers are way different. I1=-.25, I2=.166, I3=0.

Anyways my professor swears that you can use the 3 loops to solve the currents.

Me personally I don't think its possible. It looks promising and should work I guess in theory, but I've tried it in every way I can think of and can't get it to work.

Can anyone tell me why you can't use 3 loops? There must be a reason. I was think it probably had something to do with conservation of forces on the outside loop1, but i'm unsure.

Thanks for the help,
rcowboy

Okay I'm having trouble getting the image to post but heres the URL. I'll keep working on it.

http://i673.photobucket.com/albums/v...y/scan0003.jpg

Well, can't get the image to post on forum, maybe someone can help with image. Anyways link is above.
Either method will work. Check your loop 3 equation for a sign error.

4. DrRocket,

Loop3 looks good to me. Starting a point D in the lower left corner and working around clockwise I get,

-Vbd+R2I2+R3I3-Vcd=0

Thats the same I got on my paper.

5. Originally Posted by rcowboy
DrRocket,

Loop3 looks good to me. Starting a point D in the lower left corner and working around clockwise I get,

-Vbd+R2I2+R3I3-Vcd=0

Thats the same I got on my paper.
My mistake. Your loop equations are correct.

Your solution is not. The correct solution is the same as the solutions using one node equation and 3 loop equations. I1=.3, I2 =.5, I3 = .2

6. Okay, i'm totally convinced that you can't use 3 loop rules to solve this problem, or problems like it.

#1 If you were able to use three loop rules, then why did Kirchhoff come up with the junction rule?

#2 Solving the following system of equations:

-140I1+35.0I3=-35.0
-140I1-210.0I2=-147
210.0I2+35.0I3=112

I get 0=0, which means there is an infinite number of solutions. Which also means these are all identities.

I did this using the elimination method.

Either I'm doing something totally wrong, or you can't use 3 loops.

It really looks as though the 3 loops should work, but I just can't get it done.

rcowboy

7. Originally Posted by rcowboy
Okay, i'm totally convinced that you can't use 3 loop rules to solve this problem, or problems like it.

#1 If you were able to use three loop rules, then why did Kirchhoff come up with the junction rule?

#2 Solving the following system of equations:

-140I1+35.0I3=-35.0
-140I1-210.0I2=-147
210.0I2+35.0I3=112

I get 0=0, which means there is an infinite number of solutions. Which also means these are all identities.

I did this using the elimination method.

Either I'm doing something totally wrong, or you can't use 3 loops.

It really looks as though the 3 loops should work, but I just can't get it done.

rcowboy

You can most certainly use 3 loops.

You even have the equations written correctly. You just screwed up the solution. You get exactly the same solution using 3 loops as you do with 2 loops and 1 node equation.

8. Okay, check this.

Equation 1: -140I1+35.0I3=-35.0

Using elimination for the first loop means that I2 is eliminated.

the next two equations:

-140I1-210.0I2=-147
210.0I2+35.0I3=112

Eliminating I2 requires no work because they eliminate each other. yields:

-140I1+35.0I3=-35

Now I have:
-140I1+35.0I3=-35
-140I1+35.0I3=-35
-------------------------
Multiplying the top or bottom equaition by -1 yields; doesn't matter in this case. Here I did the top one.

140I1-35.0I3=35
-140I1+35.0I3=-35
-----------------------
0=0

Where am I going wrong, or should I use substitution?

rcowboy

9. Originally Posted by rcowboy
Okay, check this.

Equation 1: -140I1+35.0I3=-35.0

Using the elimination for the first loop means that I2 is eliminated.

the next two equations:

-140I1-210.0I2=-147
210.0I2+35.0I3=112

Eliminating I2 requires no work because they eliminate each other. yields:

-140I1+35.0I3=-35

Now I have:
-140I1+35.0I3=-35
-140I1+35.0I3=-35
-------------------------
Multiplying the top or bottom equaition by -1 yields; doesn't matter in this case. Here I did the top one.

140I1-35.0I3=35
-140I1+35.0I3=-35
-----------------------
0=0

Where am I going wrong, or should I use substitution?

rcowboy
I am not going to check your arithmetic. Plug in the two-loop solution. It works.

10. I am not going to check your arithmetic. Plug in the two-loop solution. It works.
lmao, thats what my professor said. Must be a teachers union thing.

11. Okay, I got confirmation from another source today that 3 loops won't work mathematically. So if someone else runs into this in the future it's noted here.

In a system of three equations, each equation must be independent of each other. In another words the outside loop (the system as a whole) is dependent on the inside loops.

If you use three loops you end up forming one of the inside loops using elimination. Just as I did in post #7. The 7th post down from the top.

So in order to solve the currents in the circuit above, you have to use 2 loop rules and 1 junction rule.

Now with that said, it only applies to the circuit above. It may not be true to other circuits in a different formation. I haven't tried other circuits yet.

rcowboy

12. Originally Posted by rcowboy
Okay, I got confirmation from another source today that 3 loops won't work mathematically. So if someone else runs into this in the future it's noted here.

In a system of three equations, each equation must be independent of each other. In another words the outside loop (the system as a whole) is dependent on the inside loops.

If you use three loops you end up forming one of the inside loops using elimination. Just as I did in post #7. The 7th post down from the top.

So in order to solve the currents in the circuit above, you have to use 2 loop rules and 1 junction rule.

Now with that said, it only applies to the circuit above. It may not be true to other circuits in a different formation. I haven't tried other circuits yet.

rcowboy
This is irrelevant.

You did not in either of your formulations use loop currents, only node currents. Whar you did was in the first case use one node equation an 2 voltage loop equations, and in the second case 3 voltage loop equations. Both approaches are valid and both yield the same solutions. You simply screwed up the solution in the second case.

13. For the record, Kirchhoff only has two rules that I know of.

#1 The sum of the currents at any junction (node), (three wires or more), =0. In another words current in equals current out.

#2 The sum of the voltages =0 for any closed loop.

Kirchhoff doesn't have a loop current rule, and I didn't use such in any of my caculations.

In post #7 I proved that 3 voltage loops doesn't work. Unless you can prove to me analytically your assumption is invalid.

rcowboy

14. Originally Posted by rcowboy
For the record, Kirchhoff only has two rules that I know of.

#1 The sum of the currents at any junction (node), (three wires or more), =0. In another words current in equals current out.

#2 The sum of the voltages =0 for any closed loop.

Kirchhoff doesn't have a loop current rule, and I didn't use such in any of my caculations.

In post #7 I proved that 3 voltage loops doesn't work. Unless you can prove to me analytically your assumption is invalid.

rcowboy
Wrong.

Kirchoff's circuit laws allow one to write down equations to solve circuit problems. They have nothing to do with the topologiy of the circuit or whether you look at loop currents or node currrents.

You can apply Kirchoff's laws with either loop currents or with nodal currents -- if you understand what you are doing. Apparently you don't.

The validity of my assertions does not in any way depend on your opoinion.

You proved absolutely nothing. The solution to your case 1 also solves case 2 (3 voltage loops), which you seem unable to understand. Moreover, your "solution" to the 3 loop formulation is wrong -- it does not satisfy the first equation unless you think that 35 = -35.

In fact, since you don't know what in the hell you are talking about, your opinion is quite irrelevant.

I have an idea. Let's ask someone with a PhD in mathematics and an MS in electrical engineering.

BTW you can solve this circuit with just 2 loop currents, which is essentially what you did in the first solution.

15. First of all the definition of Kirchhoff's loop rule is as follows: For the 2nd time.

The algebraic sum of the potential differences in any loop, including those associated with emf's and those of resistive elements, must equal zero.

I don't see anything in this definition that states anything about currents, so I don't know where your current tirade is coming from.

I will admit to a mistake I made in the three loop case. My answers should be, I1=1/4, I2=8/15 and I3=0, These are three solutions to the 3 loops, as is .3,.5,and .2. That still doesn't mean you can solve the three loops using elimination or matrices.

Since I can't get you to prove that 3 voltage loops can be solved in this way, which was my original question, your assumption is again invalid for the 2nd time.

rcowboy

16. Originally Posted by rcowboy
First of all the definition of Kirchhoff's loop rule is as follows: For the 2nd time.

The algebraic sum of the potential differences in any loop, including those associated with emf's and those of resistive elements, must equal zero.
Sonny, you need not explain Kirchoff's laws or circuit theory to me. I have done this stuff for decades. I understand it quite a bit better than do you.

Originally Posted by rcowboy
I don't see anything in this definition that states anything about currents, so I don't know where your current tirade is coming from.
That simply illustrates your lack of comprehension, not a problem with my assertion.

Originally Posted by rcowboy
I will admit to a mistake I made in the three loop case. My answers should be, I1=1/4, I2=8/15 and I3=0, These are three solutions to the 3 loops, as is .3,.5,and .2. That still doesn't mean you can solve the three loops using elimination or matrices.
Keep looking. If you wrote the equations properely you will get one and only one solution -- .3, .5 .2

Anything else is mathematically and physically impossible.

Do you honestly think that the circuit cares which equations you use to effect a solution ? Those resistors and batteries don't know or care that you exist.

Originally Posted by rcowboy
Since I can't get you to prove that 3 voltage loops can be solved in this way, which was my original question, your assumption is again invalid for the 2nd time.
My only invalid assumption was that you were a semi-intelligent fellow with a bit of understanding of elementary circuit theory. I admit to being wrong on both counts.

The classic loop current equations for your circuit are (in your notation):

Vad + Vbd= (R1 + R2) I1 + R2 I3

Vcd + Vbd = R2 I1 + (R2 + R3) I3

147 = 350 I1 + 210 I3

112 = 210 I1 + 245 I3

I1 = .3 I3 = .2

You will note that only two equations are required. To find the drop across R2 you need only note that I2 = I1 + I3

This is standard stuff for any first-year EE student.

You could also use the outer loop plus either mesh. There are several ways to skin this cat. You simply screwed up the solution to your second set of equations.

17. Well, I did learn something new today DrRocket, your simply and old hag that likes to take shots at students to boost your self-esteem. Which makes me think you couldn't cut it in the private sector and are now working for the goverment probably in the state of Wisconsin where you were protesting on a work day for your pension plan

Maybe we should go back and check your transcripts and see how well you actually did to get that PhD and MS, if thats what you got.

All you simply did was replaced R1 in the first equaition and R3 in the second with V=IR.

The same thing applies to R2I3 and R2I1

And then you use the node rule to solve for R2.

You did not solve three loop equations.

oh BTW, if this is standard stuff for a first year EE student, then thats great because you just got schooled by one pops!

I rest my case.

rcowboy

18. Originally Posted by rcowboy
Well, I did learn something new today DrRocket, your simply and old hag that likes to take shots at students to boost your self-esteem. Which makes me think you couldn't cut it in the private sector and are now working for the goverment probably in the state of Wisconsin where you were protesting on a work day for your pension plan

Maybe we should go back and check your transcripts and see how well you actually did to get that PhD and MS, if thats what you got.

All you simply did was replaced R1 in the first equaition and R3 in the second with V=IR.

The same thing applies to R2I3 and R2I1

And then you use the node rule to solve for R2.

You did not solve three loop equations.

oh BTW, if this is standard stuff for a first year EE student, then thats great because you just got schooled by one pops!

I rest my case.

rcowboy
Yopu don't have a case.

Of course the loop equations look like your firstb case. I told youbthat some time ago. You apparently finally realized that. You will need to catch on quicker if you want to become a second-year student.

You only need to write two loop equations. That is quite standard. there are several choices. Apparently you have not had a good first-year class in circuit theory. Or at least did not learn much.

Before you do any schooling you need to do some learning. A lot of learning.

The difference is that I got past, way past, first year EE classes.

You still can't find your own mistakes. Concentrate on that.

Actually I worked quite a while in the private sector. And did well enough to retire pretty early. So, you are as wrong as ever. You are clearly as proficient in being wrong as you are incompetent in circuit analysis. Thinking that formulating one problem in two different ways would result in two different solutions ought to be a hint that something is fundamentally wrong.

Now go learn what circuit analysis is all about. Maybe you need learn more about linear algebra and systems of linear equations at the same time.

19. You only need to write two loop equations. That is quite standard. there are several choices. Apparently you have not had a good first-year class in circuit theory. Or at least did not learn much.
lmao, your a trip. You can't even admit that your wrong. If I recall this whole thread began with me trying to solve 3 loop equations not 2. The question that I asked was if the circuit could be solved with 3 loops, and you said, Yes it can.

You can't solve the circuit with 3 loops as you stated in the above quote.

Evidently I have learned more then you because you still can't grasp this concept.

rcowboy

20. Originally Posted by rcowboy
You only need to write two loop equations. That is quite standard. there are several choices. Apparently you have not had a good first-year class in circuit theory. Or at least did not learn much.
lmao, your a trip. You can't even admit that your wrong. If I recall this whole thread began with me trying to solve 3 loop equations not 2. The question that I asked was if the circuit could be solved with 3 loops, and you said, Yes it can.

You can't solve the circuit with 3 loops as you stated in the above quote.

Evidently I have learned more then you because you still can't grasp this concept.

rcowboy
No stupid. If you can do it with 2 you can certainly do it with an over-determined system of 3. Any decent high school student knows that.

Or you can add a loop equation that is equivalent to I1 + I3 = I2.

You have pretty clearly learned a bunch of stuff that just isn't true.

21. No you idiot, you have to use two loop equations and one junction (node) equation.

I1+I3=I2, is a juction equation, not a loop.

22. Originally Posted by rcowboy
No you idiot, you have to use two loop equations and one junction (node) equation.
Obviously not, as I just showed you. You clearly don't know what a mesh current is. The node equation comes for free with the definition of a loop current in this case.

Go learn some circuit theory. You are making a fool of yourself -- an easy task in your case. Are you starting to get a glimpse of why your physics prof does not take you seriously ?

23. Are you starting to get a glimpse of why your physics prof does not take you seriously ?
It's funny that you said that, because today she corrected herself in front of the whole class with my name attached.

The bad part to that is now I look like the teacher pet.

24. Originally Posted by rcowboy
Are you starting to get a glimpse of why your physics prof does not take you seriously ?
It's funny that you said that, because today she corrected herself in front of the whole class with my name attached.

The bad part to that is now I look like the teacher pet.
Then your teacher needs to learn the difference between loop (mesh) currents and nodal currents. She gave in too quickly. You are still incompetent.

You can also solve your circuit problem with node equations :

92 - I1 R1 = -55 + I2 R2

57 - I3 R3 = -55 + I2 R2

I1 + I3 = I2

147 = 140 I1 + 210 I2 + 0 I3

112 = 0 I1 + 210 I2 + 35 I3

0 = I1 - I2 + I3

The unique solution is still I1 = .3 , I2 = .5 , I3 = .2

The difference is that there are three independent nodal currents but only 2 independent mesh currents. That does not stop you from writing an over-determined system of 3 loop equations which will also yield the same solution.

What you did was basically to take the nodal equations and replace 0 = I1 - I2 + I3 with the outside loop equation, and then screw up the solution of the system of equations. As noted your ( I1 = -.25, I2 = .166, I3 = 0) is not a solution. It not only fails to satisfy the first equation, yielding 35 = -35, but the idea that I3=0 is clearly wrong.

Any three linearly independent equations, properly formulated will allow a competent person to solve for I1, I2, I3. And you will get the same, unique solution.

However, you do need to be able to handle high school algebra.

25. DrRocket,

I am now beginning think you can't read.

You validated my conclusion that 3 loops won't work in your post above. Which you stated would work at the beginning of this thread

I stated in post #10 that:

In a system of three equations, each equation must be independent of each other. In another words the outside loop (the system as a whole) is dependent on the inside loops.
Since the outside loop is dependent on the inside loops, it can't be solved with three loops.

Thanks for proving that I'm right. :-D

26. Originally Posted by rcowboy
DrRocket,

I am now beginning think you can't read.

You validated my conclusion that 3 loops won't work in your post above. Which you stated would work at the beginning of this thread

I stated in post #10 that:

In a system of three equations, each equation must be independent of each other. In another words the outside loop (the system as a whole) is dependent on the inside loops.
Since the outside loop is dependent on the inside loops, it can't be solved with three loops.

Thanks for proving that I'm right. :-D
Appatently you can neither think nor read.

Since the problem can be solved with 2 loops, it can mosty certainly be solved wirth the equation for a third dependendent loop tossed in for good measure. Of course that requires someone competent in high school algebra -- which rules you out since you still can't understand your mistake.

If you really believe that I showed you to be right, and it is pretty clear that you do believe that, then we can add "delusional" to your list of problems.

I'm done wasting time on you.

27. Since the problem can be solved with 2 loops, it can mosty certainly be solved wirth the equation for a third dependendent loop tossed in for good measure.
No it can't, and you havn't proven it once.

I have proved that it didn't work in post #7

Now get out of this thread you old sore loser hag.

28. For the enlightment of lurkers and neophytes, here is rcowboy's set of "3 loop equations":

-140 I1 + 0 I2 + 35 I3 = -35

-140 I1 -210 I2 + 0 I3 = -147

0 I1 + 210 I2 + 35 I3 = 112

Now dum-dum claims that I1=-0.25, I2= 0.166, I3 = 0 solves this system of equations differing from the correct solution I1 = 0.3, I2=0.5, I3 =0.2 which you can find in my earlier posts or in rcowboy's solution by his first method, thereby showing that there isvsomething badly wrong with the "loop equations". However the fact of the matter is that his stated solutiom, I1=-0.25, I2= 0.166, I3 = 0, is NOT a solution of the given 3 equations, as almost any fool, rcowboy being a notable exception, can see. It does not satisfy even the first equation, yielding the nonsensical result 35 = -35.

On the other hand, I1 = 0.3, I2=0.5, I3 =0.2 is a solution to this system of 3 equations, and that is the correct circuit solution.

Without having to do tedious algebra one can see that I1 = 0.3, I2=0.5, I3 =0.2 is indeed a solution by direct substitution.

On the other hand

det so one can see that there are in fact infinitely many solutions.

Those solutions will be of the form I1 = 0.3, I2=0.5, I3 =0.2 plus any solution of the associated homogeneous system of equations

-140 I1 + 0 I2 + 35 I3 = 0
-140 I1 -210 I2 + 0 I3 = 0
0 I1 + 210 I2 + 35 I3 = 0

or I1 = (35/140) I3, I2 = (-35/210) I3, I3 arbitrary.

So, what is going on ?

The problem is that the equations are not in fact "loop equations" in the correct sense of linear circuit theory. In linear circuit theory the loop currents flow around their respective loops, in this case the two mesh "windows of the circuit, and the current in a common elenent, R2 for instance is the supereposition of the two loop currents. This gives the implicit relation I2 = I1 + I3 in rcowboy's notation, which adds a constraint and results in the complete system having a unique solution. So the problem is not that one cannot solve the circuit with 2 or 3 loop equations, but rather that the loop equations were not formulated properly. See my loop equation solution above for a proper formulation.

An alternative is to solve by means of nodal equations. See my subsequent post for a demonstration of how that is properly done.

So, basically rcowboy screwed up the formulation of the circuit equations, then further confused the issue by screwing up the solution of an underdetermined system of equations, failing to recognize that the correct solution was included in the solution set of his undetermined system. That should have been a clue that what was missing was an additional needed constraint. But the clue did not surface because he blew some high school algebra.

My mistake was in giving him any credit at all for understanding of circuits and being able to do basic algebra, and therefore foregoing the tedium of checking everything in gory detail, as one would for someone taking the first baby steps in circuit analysis.

You can indeed solve the circuit with just 2 loop equations -- see earlier post where this is done. You can also do it if you throw in a third loop equation, which adds nothing, BUT you have to write correct loop current equations in the first place.

What is true is that if you use nodal currents rather than true loop currents then you cannot find the necessary constraint (here I1 + I3 = I2) just from the fact that the voltage drop around any closed loop is zero. This serves as an illustration of the difference between nodal currents and loop currents.

Note also that the use of loop currents works only for circuits composed of linear elements. Kirchoff's laws apply in the non-linear case, but one is then forced to formulate the circuit description in terms of nodal currents. Superposition of loop currents is not valid in the non-linear case.

29. Ahh.. you are finally starting to make some sence.

The currents I1=-0.25, I2= 0.166, I3 = 0 are NOT solutions, and I admitted to that mistake in post #14, which either you didn't read, or can't read one.

However as I also stated in post #14 I1=1/4, I2=8/15 and I3=0 are in fact solutions to:

-140 I1 + 0 I2 + 35 I3 = -35

-140 I1 -210 I2 + 0 I3 = -147

0 I1 + 210 I2 + 35 I3 = 112

Plug them in you will see for yourself.

This means that there is a flaw using three loops which you explained greatly, I might add, in the above post.

I also showed your exact same conclusion in post #7 and I quote,

Equation 1: -140I1+35.0I3=-35.0

Using elimination for the first loop means that I2 is eliminated.

the next two equations:

-140I1-210.0I2=-147
210.0I2+35.0I3=112

Eliminating I2 requires no work because they eliminate each other. yields:

-140I1+35.0I3=-35

Now I have:
-140I1+35.0I3=-35
-140I1+35.0I3=-35
-------------------------
Multiplying the top or bottom equaition by -1 yields; doesn't matter in this case. Here I did the top one.

140I1-35.0I3=35
-140I1+35.0I3=-35
-----------------------
0=0
This means there are infinite many solutions.

Now in your post above you correct yourself and state that a constraint must be added in to form of I1 + I3 = I2, which is Kirchhoff junction rule.

I have been saying this from the very beginning of the thread and your finally catching on.

rcowboy

30. Originally Posted by rcowboy
Ahh.. you are finally starting to make some sence.

The currents I1=-0.25, I2= 0.166, I3 = 0 are NOT solutions, and I admitted to that mistake in post #14, which either you didn't read, or can't read one.

However as I also stated in post #14 I1=1/4, I2=8/15 and I3=0 are in fact solutions to:

-140 I1 + 0 I2 + 35 I3 = -35

-140 I1 -210 I2 + 0 I3 = -147

0 I1 + 210 I2 + 35 I3 = 112

Plug them in you will see for yourself.

This means that there is a flaw using three loops which you explained greatly, I might add, in the above post.

I also showed your exact same conclusion in post #7 and I quote,

Equation 1: -140I1+35.0I3=-35.0

Using elimination for the first loop means that I2 is eliminated.

the next two equations:

-140I1-210.0I2=-147
210.0I2+35.0I3=112

Eliminating I2 requires no work because they eliminate each other. yields:

-140I1+35.0I3=-35

Now I have:
-140I1+35.0I3=-35
-140I1+35.0I3=-35
-------------------------
Multiplying the top or bottom equaition by -1 yields; doesn't matter in this case. Here I did the top one.

140I1-35.0I3=35
-140I1+35.0I3=-35
-----------------------
0=0
This means there are infinite many solutions.

Now in your post above you correct yourself and state that a constraint must be added in to form of I1 + I3 = I2, which is Kirchhoff junction rule.

I have been saying this from the very beginning of the thread and your finally catching on.

rcowboy
Read more carefully. If you properly write the equations for loop currents you need only two equations. See earlier post for just such a solution.

If you insist on a nodal formulation then you require 3 equations.

There is more than one way to skin this cat. You have not understood the more efficient, for hand calculations, loop current approach.

Your problem, besides multiple mistakes, lies in mixing the two methods and not understanding the distinction. To compopund the communication problem, you use standard terminology with non-standard meanings.

Nodal equations are easily automated for a description of te circuit in terms of graph theory, and are therefore often used in circuit analysis programs -- but the initial form is a bit different, though equivalent to yours. See my earlier post for an example.

31. To further your education with respect to loop currents here are two other ways to formulate equations for your circuit, this time using the outer loop. All loop currents are chosen to flow clockwise.

I. Let be the outer loop current and be the current in the first mesh. In your notation this means

, and

Then the loop equations are

II. Let be the outer loop current and be the current in the second mesh. In your notation this means

, and

Then the loop equations are

Note that the "outer loop current" is different in these two cases. The definition of loop currents depends on the entire set of loop currents -- loop currents have no individual identity. Loop currents are in that regard more abstract than are nodal currents.

In each case there will be a unique solution and it will completely determine all currents and all voltage drops in the circuit.

The general tereminology in circuit analysis is that "loop equations" refer to a loop current formulation and "nodal equations" refer to a nodal formulation. Physical current and voltage measurements reflect nodal currents, which in general are superpositions of loop currents.

These distinctions are standard in electrical engineering and very important in the use and development of circuit analysis software. For more detail see, for instance The Analysis of Linear Circuits by Close.

32. DrRocket,

That alittle more in-depth then I was looking for lol. I appericate it.

I understand that its inefficient using three equations, and that was basically my whole point.

I would like for you do to a small favor for me tough. A lot of people come here looking for help on subject matters that are sometimes, well, not explained very good in books or in class rooms. They are looking for alternative sources to help supplement or fill in the voids if you will.

When you come down on them it does two things:

1. Discourages people from using the forum.

2. Causes people to lose interest in the field of study that your talking about, and for the field of EE were short on people as it is.

Of course I'm hard headed, and you can say what ever you want to me.

Anyways, just keep this in mind for me.

rcowboy

33. Originally Posted by rcowboy
DrRocket,

That alittle more in-depth then I was looking for lol. I appericate it.

I understand that its inefficient using three equations, and that was basically my whole point.

I would like for you do to a small favor for me tough. A lot of people come here looking for help on subject matters that are sometimes, well, not explained very good in books or in class rooms. They are looking for alternative sources to help supplement or fill in the voids if you will.

When you come down on them it does two things:

1. Discourages people from using the forum.

2. Causes people to lose interest in the field of study that your talking about, and for the field of EE were short on people as it is.

Of course I'm hard headed, and you can say what ever you want to me.

Anyways, just keep this in mind for me.

rcowboy

Tell the person helping that nhe is wrong and his "assumptions are invalid" and you get stringly corrected. If you can't stand the heat, stay out of the kitchen.

34. Originally Posted by DrRocket
For the enlightment of lurkers and neophytes, here is rcowboy's set of "3 loop equations":

-140 I1 + 0 I2 + 35 I3 = -35

-140 I1 -210 I2 + 0 I3 = -147

0 I1 + 210 I2 + 35 I3 = 112

Now dum-dum claims that I1=-0.25, I2= 0.166, I3 = 0 solves this system of equations differing from the correct solution I1 = 0.3, I2=0.5, I3 =0.2 which you can find in my earlier posts or in rcowboy's solution by his first method, thereby showing that there isvsomething badly wrong with the "loop equations". However the fact of the matter is that his stated solutiom, I1=-0.25, I2= 0.166, I3 = 0, is NOT a solution of the given 3 equations, as almost any fool, rcowboy being a notable exception, can see. It does not satisfy even the first equation, yielding the nonsensical result 35 = -35.

On the other hand, I1 = 0.3, I2=0.5, I3 =0.2 is a solution to this system of 3 equations, and that is the correct circuit solution.

Without having to do tedious algebra one can see that I1 = 0.3, I2=0.5, I3 =0.2 is indeed a solution by direct substitution.

On the other hand

det so one can see that there are in fact infinitely many solutions.

Those solutions will be of the form I1 = 0.3, I2=0.5, I3 =0.2 plus any solution of the associated homogeneous system of equations

-140 I1 + 0 I2 + 35 I3 = 0
-140 I1 -210 I2 + 0 I3 = 0
0 I1 + 210 I2 + 35 I3 = 0

or I1 = (35/140) I3, I2 = (-35/210) I3, I3 arbitrary.

So, what is going on ?

The problem is that the equations are not in fact "loop equations" in the correct sense of linear circuit theory. In linear circuit theory the loop currents flow around their respective loops, in this case the two mesh "windows of the circuit, and the current in a common elenent, R2 for instance is the supereposition of the two loop currents. This gives the implicit relation I2 = I1 + I3 in rcowboy's notation, which adds a constraint and results in the complete system having a unique solution. So the problem is not that one cannot solve the circuit with 2 or 3 loop equations, but rather that the loop equations were not formulated properly. See my loop equation solution above for a proper formulation.

An alternative is to solve by means of nodal equations. See my subsequent post for a demonstration of how that is properly done.

So, basically rcowboy screwed up the formulation of the circuit equations, then further confused the issue by screwing up the solution of an underdetermined system of equations, failing to recognize that the correct solution was included in the solution set of his undetermined system. That should have been a clue that what was missing was an additional needed constraint. But the clue did not surface because he blew some high school algebra.

My mistake was in giving him any credit at all for understanding of circuits and being able to do basic algebra, and therefore foregoing the tedium of checking everything in gory detail, as one would for someone taking the first baby steps in circuit analysis.

You can indeed solve the circuit with just 2 loop equations -- see earlier post where this is done. You can also do it if you throw in a third loop equation, which adds nothing, BUT you have to write correct loop current equations in the first place.

What is true is that if you use nodal currents rather than true loop currents then you cannot find the necessary constraint (here I1 + I3 = I2) just from the fact that the voltage drop around any closed loop is zero. This serves as an illustration of the difference between nodal currents and loop currents.

Note also that the use of loop currents works only for circuits composed of linear elements. Kirchoff's laws apply in the non-linear case, but one is then forced to formulate the circuit description in terms of nodal currents. Superposition of loop currents is not valid in the non-linear case.
Here is alittle typo in the 3 times 3 determinant calculation, its 210, not -210.

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