Kcl circuit analysis problem

[Daniel Wood]

Hi guys,

I have been trying to keep on top of what I learnt at college and after many attempts, I just cant work out the currents in each resistor on the circuit shown below.

I think the confusion comes from the direction of current in each component. This ultimately leads to the calculations being different from the simulation. So anyway I assume that the current flow will be clockwise on both loops.

This is what i thought the analysis should be following the paths of E1 and E2. But we know its wrong
[LATEX]\epsilon _1 = -9 + 4_{i1} +12(i_1 - i_2) \newline \epsilon _2 = -6 + 8_{i2} +12(i_1 - i_2) [/LATEX]
After a re-arrangement we get:
[LATEX] \binom{16_{i1} - 12_{i2} = 9}{12_{i1} - 4_{i2} = 6} [/LATEX]
But we know it should be this to get the correct answers
[LATEX] \binom{16_{i1} - 12_{i2} = 9}{12_{i1} - 20_{i2} = -6} [/LATEX]

Can someone please provide some insight on how to get the correct analysis of E1 and E2 without using trial and error?
Thanks

 

[steveB]

yeah, in the second equation, the second term for the 12 ohm resistor should be i2-i1. This is needed to get the polarity correct on the voltage drop.

 

[Ian Rogers]

I'm sitting here with Danny.... We have been going mad.... All the info on the net talk about I1 = I2 + I3

So we assumed I3 is equal to I1 - I2... The trouble here is that all three currents are positive ( as drawn ) so I personally assumed 0 = +6 + 8*i2 + 12(i1 - i2)... Should it have been 0 = +6 +8*i2 + 12(i2 - i1)

Which yields -12*i1 + 20*i2 = 6 ..

 

[steveB]

Those last two equations don't match. 0 = +6 +8*i2 + 12(i2 - i1) is not the same as -12*i1 + 20*i2 = 6

The latter one is correct.

 

[Ian Rogers]

Those last two equations don't match. 0 = +6 +8*i2 + 12(i2 - i1) is not the same as -12*i1 + 20*i2 = 6

The latter one is correct.

My bad... I keep putting the signs down wrong... The confusion in the first place is because ALL example on the net have the batteries ( voltage sources ) opposite so you can reference 0v at the bottom node... This is why I keep writing it down wrong.... I find the only way to get this right is to do a clockwise loop for i1 and an anticlockwise loop for i2 Then I seem to write the signs down correctly....( I did this over 25 years ago.... My brain hurts..)

 

[steveB]

Yes, I know the feeling. We get used to seeing things a certain way and a flip, turn or mirror image in a schematic can throw a wrench in our thinking. The best work around for this is to just develop your own rule set that you follow and then just blindly follow it while you write out the equations. No matter how much experience you have, it's still easy to make a mistake in signs.

 

[Ian Rogers]

Trouble here is.... I stick it in ISIS and the results are instant... I can see why folks don't learn this anymore..
We watched a video on Kirchoffs law.... I was totally lost... The interpretation of which way the current flows is mind boggling... First thought you CAN'T see which way they are flowing.... Is it just guess work

 

[MrAl]

Hi,

I think this has already been resolved, but i think if the loops had been labeled "i1" and "i2" in the first place then the polarities in the second equation would have been correct because it would have been clear that i2 was the current of interest in the second loop so that one is positive. Maybe that would have helped anyway.

So we have:

[LATEX]-9 + 4_{i1} +12(i_1 - i_2)=0 \newline -6 + 8_{i2} +12(i_2 - i_1) =0[/LATEX]

Also, i would suggest looking into nodal analysis:

[LATEX]\frac{V1-v}{R1}-\frac{v}{R2}-\frac{v-V2}{R3}=0[/LATEX]

where v is the center voltage and v is the only unknown (V1=9, V2=-6). Note this works well with voltage sources.

Ian:
In many of these problems the current direction is "assumed"
BTW what kind of beer is that in your avatar? I might want to try it, i've tried many different beers from around the world.

 

[steveB]

.... Is it just guess work

Guesswork will make half the signs wrong, but even bad students do better than that. (joking)

Seriously though, you can work out a procedure, and each person might do it in their own way, but there are basic rules to follow. In another forum, I once wrote out my way. I'll attach it here.

With mesh currents, as you were using, you have to think a little bit more, but the basic rules still apply.

 

[Ian Rogers]

Here's an excellent example.... In your PDF R2 is shown positive towards the bottom of the circuit.... Here would be my first mistake I would have assumed the opposite..

 

[MrAl]

Hi Steve,

Yes that's funny, guesswork will make 50 percent right and 50 percent wrong on average, so we could set up a procedure to gradually correct all incorrect polarities with trial and error

If we always make the currents loop clockwise (or all counter clockwise) i think we're ok.

 

[steveB]

Wait,

I just realized I should ask and make sure. Were you trying to use mesh current analysis?

 

[MrAl]

Hi Steve,

Who are you asking ?

 

[Ian Rogers]

Its all down to I 3.... The 12 ohm resistor.... If I go clockwise on both loops I must remember to note the current direction change.... I sometimes how I passed this sooo long ago.... Couldn't do it anymore.... Most of my design work nowadays is digital...

 

[Ian Rogers]

Wait,

I just realized I should ask and make sure. Were you trying to use mesh current analysis?

Yes!!

 

[steveB]

Hi MrAl,

I was asking Ian and Daniel. I understood your post, and was pretty sure I understood theirs too, but just wanted to verify.

Thanks everyone, ... seems like it is resolved now.

 

[Ratchit]

Daviel Wood,

"I think the confusion comes from the direction of current in each component. This ultimately leads to the calculations being different from the simulation. So anyway I assume that the current flow will be clockwise on both loops."

Yes, that is the best way to do it. I believe steveB showed you where your mistake was. Always assume your mathematical current is CW even if there is a physical current source in the opposite direction. If the computed mathematical value is negative, that shows your physical current is opposite of your mathematical assumption (CW). E1,E1,epsilon1, and epsilon2 are all zero and a distraction.

Ian Rogers,

"I'm sitting here with Danny.... We have been going mad.... All the info on the net talk about I1 = I2 + I3
So we assumed I3 is equal to I1 - I2... The trouble here is that all three currents are positive"

Why three currents for two loops? Just extra work and confusion.

"I find the only way to get this right is to do a clockwise loop for i1 and an anticlockwise loop for i2"

Wrong! You are going to get wrapped around the axle doing that. You will get the same answer doing a KVL either CW or CCW, but be consistent with each loop. You will also find it is much easier to figure out the voltages across components common to two loops than splitting the currents between two or more loops (it's bad to do I3 = I1 - I2).

"Trouble here is.... I stick it in ISIS and the results are instant... I can see why folks don't learn this anymore.."

What is ISIS? One of the definitions in the acronym dictionary is 'Internet Sexuality Information Services'. Which definition did you have in mind?

"We watched a video on Kirchoffs law.... I was totally lost... The interpretation of which way the current flows is mind boggling... First thought you CAN'T see which way they are flowing.... Is it just guess work"

As I said above, assume CW current., calculate, then determine the physical direction from the calculation results and asssumption.

MrAl,

I believe the term (v-V2)/R3 your node equation should be (v+V2)/R3, because V2 is increasing the current out of the node.

steveB,

"Seriously though, you can work out a procedure, and each person might do it in their own way, but there are basic rules to follow. In another forum, I once wrote out my way. I'll attach it here."

I am confused by your example. Why are there three currents for two loops and two equations? What is the difference between the loop2 current and I3? I would never assign polarity to passive conponents. I would let the assumed currents define what the voltage across them should be.

Ratch

 

[steveB]

steveB,

"Seriously though, you can work out a procedure, and each person might do it in their own way, but there are basic rules to follow. In another forum, I once wrote out my way. I'll attach it here."

I am confused by your example. Why are there three currents for two loops and two equations? What is the difference between the loop2 current and I3? I would never assign polarity to passive conponents. I would let the assumed currents define what the voltage across them should be.

Ratch

Yes, it is confusing. Unfortunately, that's an example I did for someone at Physics Forums where they were not using mesh current analysis but were just applying KVL directly. The mesh current approach is simpler in these cases, I think. So, the rules I showed (which are not in any way unique) are helpful for setting up the problem and getting the signs right when doing KVL loop analysis. A little more thought is needed to make sure the mesh equations are done consistently. Still, the lesson here is that we don't need to use guesswork if we develop a systematic procedure ahead of time and stick to it.

 

[Ian Rogers]

Wrong! You are going to get wrapped around the axle doing that. You will get the same answer doing a KVL either CW or CCW, but be consistent with each loop. You will also find it is much easier to figure out the voltages across components common to two loops than splitting the currents between two or more loops (it's bad to do I3 = I1 - I2).

Not really!!! It works that way round... But all hail Steve for pointing out that I1 - I2 does indeed = I3. Similarly I2 - I1 = inverse I3... I just didn't think it through..... If Danny brings in another one I'll shoot him... I am grateful to all.. Thank a bundle...

 

[MrAl]

MrAl,

I believe the term (v-V2)/R3 your node equation should be (v+V2)/R3, because V2 is increasing the current out of the node.

Ratch

Hello there Ratch,

Well, i wrote the equation for a more general network where i write based on the topology alone, then fill in the blanks later. This allows me to use the same equation for any circuit no matter what the polarity of the voltage sources are. So the equation is correct as it is, but you have to know how to apply it. This would be the case even if we did it the way you suggest by assuming right off that the right hand side voltage is actually negative to start with, but i prefer to do it with unassigned polarities and then fill them in later. This also helps relate the variables to the voltages without assuming anything so it is more general i think that way.

For example, if we keep the sign the way it is and we had a program where we fill in the voltage for V1 and V2 we would have two lines that look like this:
V1=9
V2=-6

and that makes sense because V1 has it's negative terminal at ground so it is a positive source and V2 has it's positive terminal at ground so it's a negative source. If we wanted to change V2 to a positive source we could just do:
V1=9
V2=6

and the equation would still work correctly.

If we instead change the sign as you suggested (which should lead to the same result if applied correctly) with the V2 as a negative source we would have in our program:
V1=9
V2=6

and if later we wished to change the V2 source to a positive source would have to change that to:
V1=9
V2=-6

So you see how this could be confusing, where we have to enter a negative number in for a positive source in order to get the correct results with the *same* equation (more generally the same set of equations).

See the attached diagram where this becomes a little more clear. You can see that the topology does not assume anything but the connections between the components themselves, and that's all that a topology should do anyway.
Recall that in nodal analysis we write the equations for the node voltages referenced to ground not for the actual source voltages. Doing this follows almost the same principle as always assuming a clockwise current, because we may not know what it really is yet so we do something that can be done in any circuit.