# Mesh Current question

Discussion in 'Homework Help' started by Curls, Jan 9, 2010.

1. ### Curls Thread Starter New Member

Jan 6, 2010
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Hello again,

since help came so quickly with my previous question, I thought I'd give it a new shot
Here http://www.allaboutcircuits.com/worksheets/dcmesh.html I don't understand question 4.

The way I see it: current in the first loop goes clockwise, so polarities of resistors are:
The left-most resistor: left -, right + (since current upstream/enters resistor, so negative on the left side, downstream/exit resistor, positive on the right side). Correct?
Middle resistor (as seen from the first loop mesh current): top -, bottom + (same reason). Is this correct?

In that case, when you'd imaginary place a voltmeter over all the loads and go around the first loop and start at the 6V battery with red lead at the positive terminal and black lead at the negative, you'd get a reading of 6, then I'd move the imaginary meter up over the left most resistor, I'd get a reading of 1000I1 Volts, then to the middle resistor, a reading of 1000(I1-I2) Volts, then on the 1 volt source of course a reading of 1V. These figures would ALL be positive, since the red lead would consequently be attached to a positive terminal (both resistors AND batteries have their polarities in the same, assumed Mesh-current direction, right?)

So how come the KVL equitation given in the answers is: 6 − 1000I1 − 1000(I1 + I2) + 1 = 0 ??

I'm probably seeing this wrong but I compared it to other exercises and to the theory described in the chapter and I can't seem to find what's wrong in my way of thinking here...

Thanks!

S.

Feb 17, 2009
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3. ### Curls Thread Starter New Member

Jan 6, 2010
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Hi! Thanks for the drawing It still isn't completely clear to me though :s Try to tell me, if possible, what's wrong with my reasoning: the supposed Mesh current travels in the left most resistor from left to right, negative left, positive right; to create my KVL equitation, I'd start metering from the 6 voltage source, with red lead at the top (positive pole) and black lead at the bottom, then take voltmeter, place it over the load, red lead at the right side (positive (mesh current exits)), black lead left (negative (mesh current enters), so I'd get a positive voltage... No? Bit confused, sorry...

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If I understand you correct voltage will be negative because you assume that current flow from + to - . Of course you can choose whatever direction you wish for any loop.
And it doesn't matter which you choose

For V3=6V and V2=0.6V and clockwise direction
V1+V2-V3=0
then
V1=V3-V2=5.4V

And for counter clockwise

V3-V1-V2=0V or
(-V1)+(-V2)+V3=0V
(-5.4V)+(-0.6V)+6V=0

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5. ### Curls Thread Starter New Member

Jan 6, 2010
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I really apreciate your efforts to try to explain it to me but the more I read about it and the more I try to follow your answers, the more confused I get... It just doesn't make sense when I compare it to the theory explained here http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

I don't assume that current flows from + to -, but the above theory states the following:

The next step is to label all voltage drop polarities across resistors according to the assumed directions of the mesh currents. Remember that the upstream end of a resistor will always be negative, and the downstream end of a resistor positive with respect to each other, since electrons are negatively charged. The battery polarities, of course, are dictated by their symbol orientations in the diagram, and may or may not agree with the resistor polarities (assumed current directions)

So that's exactly what I do, but when I look at exercises and other examples online, it simply doesn't seem to work. For example, aside from the previous question; question 7 on the same page gives the same problem:

If I follow the outer loop I1, according to the theory above, the polarity of resistor R2 should be negative on the left, positive on the right, (mess current entering and exiting), then entering resistor R5, so negative on the right, positive on the left. So both resistors (R2 and R5) AND battery in this loop have the same polarity direction, so the KVL should be: 10 + 50(I1+I2) + 120(I1-I3) = 0, which would give 10 + 170I1 + 50I2 - 120I3 = 0 or 170I1+50I2-120I3 = -10, but as you can see in the answers it says 170I1 + 50I2 -120I3 = 10 (instead of -10)...

The way I do it here is EXACTLY the same as how it is described in the theory example here: . In I3-loop, all the polarities are in the same direction if you follow the mesh current, so the equitation is ...+...+...=0, just like I do in the exercises above ...

So, can anyone explain to me what would be the correct way to determine the polarities and how they should be added up, because if I do it as explained in the theory, it doesn't work...

Once again, thanks for your efforts, but as I said: the more I read about it, the more confused I get...

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Maybe you solve this circuit in "your" way

And later on with this equations
6 − 1000I1 − 1000(I1 + I2) + 1 = 0
7.2 − 1000I2 − 1000(I1 + I2) + 1 = 0
And then we compare the results.
And remember if current is negative this means that is flow in opposite direction that you assume.
and then you see it doesn't matter which method you use

7. ### Curls Thread Starter New Member

Jan 6, 2010
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Ok I dit it according to my equations (so with all terms positive as described above) and I end up with I1=-0,00193A and I2=-0,00313A, so both assumed Mesh current directions are "wrong" and current flows opposite...
But if you work out the "correct" or other equations, you end up with +0,00193A and +0,00313A, so according to these, the Mesh currents are right? How do you explain that?

Last edited: Jan 10, 2010
8. ### Curls Thread Starter New Member

Jan 6, 2010
18
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Hmm, after a few extra exercises where I keep getting exact opposite results (in + and -), let me guess: just a matter of conventional versus real current flow?

The way I get it and draw my answers, I "draw" real current flow answers (from - to +), but if I make drawings with the "other/correct" answers, current flows conventional, from + to -. Am I thinking right here? In that case: question/problem solved
And also in that case, one minor detail though: I think it's a little bit confusing if the author of the theory describes a whole theory following "real" current flow and then works out his exercises with conventional current flow... As I said before, question 7 here http://www.allaboutcircuits.com/worksheets/dcmesh.html has the same thing. If you work out his "correct" equations only I2 has a negative current so the flow of I1 and I3 are correct and you have to change I2. Conventionally seen, that's right, but if you create equations according to what's explained in the theory, you get I1 and I3 negative and I2 ok, so current flows "real", from - tot +... Confusing!

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But what is there to explain
In your method you assume that current is flow in this direction

So when you get "negative" current that means that current is flow in opposite direction than you assume.
In theory current flow from + to -.

And in question 7 they assume that current is flow in this direction.
The tip (pike) of a arrow show "positive" voltage .

And yes you right it is a bit confusing that I always analyse circuit with conventional current flow is from + to -.

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10. ### Curls Thread Starter New Member

Jan 6, 2010
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What is there to explain? Ok, one last time.

Before I studied the theory of Mesh currents on all about circuits, I knew nothing about it, so I can only speak with the knowledge I got from this site.

I'm going to take question 7 here http://www.allaboutcircuits.com/worksheets/dcmesh.html as example. When you work out the "correct" equations, I1 and I3 are positive so assumed Mesh current flow is OK, I2 is negative, so assumed Mesh current is wrong, so THIS is the right answer: , blue arrows indicate current flow.

According to MY equations, which I created using the theory described on this site, I1 and I3 are negative so assumed Mesh current flow is wrong, I2 is positive, so assumed Mesh current is ok, so THIS is the right answer:
, blue arrows indicate current flow.

When you compare my answer to the almost EXACT same example given in the theory, you get the same thing, current flows from - to +: .

So the only thing I'm saying is that it's quiet confusing to read an example in the theory using REAL current flow (from - to +) and when you then try to solve an exercise based on that theory you just read, current flow is suddenly conventional...

Pfew Hope I'm finally making myself clear about this and hope I understand it at last Thanks for your answers, they were helpful!

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well I understand you pint, bur hey what can I do about that.
You just need to remember that this hole "electric theory" was invented for conventional current flow.

12. ### Curls Thread Starter New Member

Jan 6, 2010
18
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Haha, you can't, of course, do anything about that, that was not the point At first I didn't understand the mesh-thing, then with your help I did and THEN I just noticed that the difference between theory/exercises on the same site is a bit confusing; but that's nobody's fault, that's just... a fact

Thanks a lot for your help!

13. ### hitmen Active Member

Sep 21, 2008
159
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Yes if the current is "wrong". It means that your assumed direction is wrong. Have you learnt about the passive sign convention?

14. ### Curls Thread Starter New Member

Jan 6, 2010
18
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Ehm, thought I had it, but your comment makes me doubt again If the current is "wrong" (in this case wrong is "negative"), that means that the assumed direction is wrong, yes, obviously, that's how it is explained in the theory, so that's how I understand it...
Didn't read anything about a passive sign convention though...

15. ### Curls Thread Starter New Member

Jan 6, 2010
18
0
Ok, found it somewhere on the internet:

Here are some basic ground rules:

• All resistors are either positive or negative uniformly. Which means that if you consider one resistor to be positive (which is the common case) then all the resistors are positive.

• At least one source is the opposite sign of the resistors. If only one is present then that is the one.

• Always start by making your loop.
Now the correct equations given in the answers to the Mesh Current questions on this site begin to make sense, but once again and once and for all , that's NOT how it is explained and described in the Mesh Current Theory on this site... Look at the examples in the theory, sometimes ALL of the components in one loop (resistors AND voltage source) are ALL positive (which goes against the second point in the above passive component convention rules...)

But it's no longer a problem to me as I've noticed that my method (which is the same as the AAC theory) works well... And with this passive sign convention thing I understand now also how the equations in the exercises are made up.

Just one question though, for future development and other analysis: is one theory preferable over the other?

Thanks for the help...