*reposted in homework help forum. Sorry bout posting it in the wrong place earlier.
I cannot figure out how to do this. It is an example problem for test prep. We have to use Kirchoff's rules to find three equations, then solve using matrices, etc.
Here is a link to a picture of the circuit.
http://img375.imageshack.us/my.php?image=k...ecircuitoj6.png
Given the above:
( a ) Applying Kirchoff's voltage and current rules write three sets of equations
( b ) Put the 3 equations using matrix representation
( c ) Find the determinant of the coefficient matrix
( d ) Find the cofactor matrix for the coefficient matrix
( e ) Find the inverse matrix for the coefficient matrix
( f ) Applying the inverse matrix, find the current i1, i2, and i3.
For c through e I am pretty sure I can handle it. But I can't figure out the easiest part (allegedly), part (a), and the last part might I'm not sure, but I DO know how to find the inverse matrix (just take your matrix, put the identity matrix beside it, forming one "supermatrix," and then row reduce it).
But basically, if I can't get past part (a) I can't do any of it. Here is what I have gotten so far:
Using the voltage rule:
Loop 1:
-V1 - i1R1 + V2 + i2R2 -i1R1 = 0 ==> -V1 + V2 -2i1R1 + i2R2 =0
Loop 2:
-i2R2 - V2 - i3R1 + V2 - I3R1 = 0 ==> -i2R2 - 2i3R1 = 0
Loop 3: This one just gives the same thing as Loop 1.
Using the current rule:
i3 - i1 - i2 = 0
I imagine I am missing something terribly obvious, but I am confused because I have several terms of just i's and several with iR's. So, do I have the equations right? And, if I do, how do I write them in a useful matrix form?
Thanks yet again!
Hi Grasshopper,
Check your loop 1
V1 <> V2
Loop3? Two should be enough.
-C2.
Edit .. use your own I1 and I2 for each loop .. they're just trying to confuse you with theirs
Check your loop 1
V1 <> V2
Loop3? Two should be enough.
-C2.
Edit .. use your own I1 and I2 for each loop .. they're just trying to confuse you with theirs
Are you saying that I can just assume that V1 is interchangeable with V2? (that is, I can treat them as equal to each other?) Cause if so, I can see where that would seriously simplify things. But I'm not sure if that's what you mean.
Actually, I'm going to go ahead and make a similar assumption. What if I do this?
-V1 + V2 -2i1R1 + i2R2 =0
where -V1 + V2 = some voltage ∆V
Giving
-2i1R1 + i2R2 = ∆V
since I'm dealing with coeficient matrices only, could I do this?
-i1 + i2 = 1
And then with loop 2 I have:
-i2R2 - 2i3R1 = 0
doing the same thing gives:
-i2 + i3 = 0
And finally, with the current rule:
i3 - i1 - i2 = 0
I will rewrite all of them with in order, giving:
-i1 + i2 + 0i3 = 1
0i1 - i2 + i3 = 0
-i1 -i2 + i3 = 0
Then I can write them in a coefficient matrix, giving
| -1 1 0|
|0 -1 1|
|-1 -2 1|
Am I still way off? To be honest I'm not really seeing how I can pull all this. It seems kind of arbitrary and made up. :shrugs:
But anyway, and then I have:
| -1 1 0| | i1| = |1|
|0 -1 1| |i2| = |0|
|-1 -2 1| |i3| = | 0|
* by the way, how do you code a matrix here? Anyone want to take the time to explain?
-V1 + V2 -2i1R1 + i2R2 =0
where -V1 + V2 = some voltage ∆V
Giving
-2i1R1 + i2R2 = ∆V
since I'm dealing with coeficient matrices only, could I do this?
-i1 + i2 = 1
And then with loop 2 I have:
-i2R2 - 2i3R1 = 0
doing the same thing gives:
-i2 + i3 = 0
And finally, with the current rule:
i3 - i1 - i2 = 0
I will rewrite all of them with in order, giving:
-i1 + i2 + 0i3 = 1
0i1 - i2 + i3 = 0
-i1 -i2 + i3 = 0
Then I can write them in a coefficient matrix, giving
| -1 1 0|
|0 -1 1|
|-1 -2 1|
Am I still way off? To be honest I'm not really seeing how I can pull all this. It seems kind of arbitrary and made up. :shrugs:
But anyway, and then I have:
| -1 1 0| | i1| = |1|
|0 -1 1| |i2| = |0|
|-1 -2 1| |i3| = | 0|
* by the way, how do you code a matrix here? Anyone want to take the time to explain?
I can't open the link at work so no help with the easy bit yet 
CODE
Try [ code ], [ /code ] tags (without the spaces) and preview your post to get the lining-up right, that should do the trick for matrices.
QUOTE (bm1957+Oct 20 2008, 04:17 PM)
I can't open the link at work so no help with the easy bit yet
CODE
Try [ code ], [ /code ] tags (without the spaces) and preview your post to get the lining-up right, that should do the trick for matrices.
Testing matrix code thing you told me:
CODE
|-1 1 0|
| |
|0 -1 1|
| |
|-1-2 1|
| |
|0 -1 1|
| |
|-1-2 1|
Seems about right.
Okay, I will just manually write the circuit in text using the code thing. Hopefully this works. Basically it's a circuit with two loops connected by a wire down the middle. V1, V2, i1, i2, i3, R1 and R2 make up the circuit. The current basically flows left to right on top, right to left on bottom, and down the middle wire.
EDIT- Okay, that didn't work. Gonna try again.
Forget the code. i can't get it. Here is the circuit in text (ignore .......)
_____R1___(i1 goes <-- here)_____________( i3 goes <--)______
| .......................................................|..........................................................|
|........................................................|.........................................................V2
V1...............(i2 goes down here)R2........................................................|
|........................................................|............................................................|
|.......................................................V2........................................................R1
|........................................................|............................................................|
|........................................................|............................................................|
|___R1__ (i1 goes --> here)__..|___________(i3 goes -->)______|
Hi Grasshopper,Bm1957,
Sorry, my first post was made just as I was leaving for work .. I didn't read the question .. not good! Also,in my fevered state <> was supposed to mean 'not equal' rather than 'interchangeable'. FWIW I use a 'method' (mesh analysis) to deal with these things .. clearly your teacher wants you to have to think. I'm forced to work through the approach the questioner intends to see if I get the same result as you. Time passes ..
Meanwhile I suspect the 'trick' (extra equation) is to see that
i2 = i3 - i1
-C2.
Edit .. Your later post wasn't in my cache .. I get the same voltages going round the loops as you do. But your -i1 + i2 = 1 doesn't look good to me .
Sorry, my first post was made just as I was leaving for work .. I didn't read the question .. not good! Also,in my fevered state <> was supposed to mean 'not equal' rather than 'interchangeable'. FWIW I use a 'method' (mesh analysis) to deal with these things .. clearly your teacher wants you to have to think. I'm forced to work through the approach the questioner intends to see if I get the same result as you. Time passes ..
Meanwhile I suspect the 'trick' (extra equation) is to see that
i2 = i3 - i1
-C2.
Edit .. Your later post wasn't in my cache .. I get the same voltages going round the loops as you do. But your -i1 + i2 = 1 doesn't look good to me
.
C2, you are a holdout. 
The Joint is waiting, and your company is very welcome.
Y'all can use LaTeX or upload images just like a real discussion board.
The Joint is waiting, and your company is very welcome.
Y'all can use LaTeX or upload images just like a real discussion board.
Interestingly, on the test we had a circuit with actual values for voltage and resistence. I guess he felt sorry for us. Still, I need to get better at this one.
---------------------------------------------------------------------------------------
Sapo, I think your link is dead.
---------------------------------------------------------------------------------------
Sapo, I think your link is dead.
I think that Loop 1 and Loop 3 are different. The sign on V2 will be different (I think), and you get a -i2R1 term, not +i2R2
Three loops, 3 equations, eliminate i3 like you did before.
Been a while since I did this but I remember it really helped to draw loops on the diagram and to draw your assumptions on directions of voltage at voltage sources to help remember your signs.
Hopefully that will help some?
The joint
Three loops, 3 equations, eliminate i3 like you did before.
Been a while since I did this but I remember it really helped to draw loops on the diagram and to draw your assumptions on directions of voltage at voltage sources to help remember your signs.
Hopefully that will help some?
The joint
I confirm that sensible people would take two loop currents as inverting a 2*2 matrix is easier.
In fact, there are more simple methods then (called fluence graphs, at least in French) to avoid any matrix inversion.
But if you can't change you teacher, then write that voltages across the 3 paths are equal:
V1 + 2*R1*i1 = V2 + R2*i2 = V2 - 2*R1*i3
and that i3=i1+i2
Good luck!
In fact, there are more simple methods then (called fluence graphs, at least in French) to avoid any matrix inversion.
But if you can't change you teacher, then write that voltages across the 3 paths are equal:
V1 + 2*R1*i1 = V2 + R2*i2 = V2 - 2*R1*i3
and that i3=i1+i2
Good luck!
QUOTE (Enthalpy+)
In fact, there are more simple methods then (called fluence graphs, at least in French) to avoid any matrix inversion.
I wasn't planning to do any matrix inversion (whatever that be) either. Show me yours?
-C2.
I wasn't planning to do any matrix inversion (whatever that be) either. Show me yours?
-C2.
QUOTE (Enthalpy+Oct 22 2008, 12:11 AM)
I confirm that sensible people would take two loop currents as inverting a 2*2 matrix is easier.
But this circumvents the fact that an introduction to this subject teaches 3 loops, one for each half of the circuit and one for the outside loop. Kirchoff's voltage law is applied to each loop (as implied by the question) and Kirchoff's current law is used to eliminate some variable(s).
There are certainly better methods but jumping into Einstein's equations as an introduction to GR wouldn't be sensible, and neither would be trying to skip introductory methods in this subject.
But this circumvents the fact that an introduction to this subject teaches 3 loops, one for each half of the circuit and one for the outside loop. Kirchoff's voltage law is applied to each loop (as implied by the question) and Kirchoff's current law is used to eliminate some variable(s).
There are certainly better methods but jumping into Einstein's equations as an introduction to GR wouldn't be sensible, and neither would be trying to skip introductory methods in this subject.
The question already uses i3 so an outer loop current would have to be i4. The teacher might reasonably ask why the student had skipped i3 .. which appears in the question. The student might choose to give an answer using i1,i2,i3 and i4 .. but I suspect this could not be solved with three equations and the question clearly expects three equations to be used.
Hopefully Grasshopper will be able to reveal how the teacher intended the question to be answered.
-C2
Hopefully Grasshopper will be able to reveal how the teacher intended the question to be answered.
-C2
QUOTE (Confused2+Oct 22 2008, 08:34 PM)
The question already uses i3 so an outer loop current would have to be i4. The teacher might reasonably ask why the student had skipped i3 .. which appears in the question. The student might choose to give an answer using i1,i2,i3 and i4 .. but I suspect this could not be solved with three equations and the question clearly expects three equations to be used.
Hopefully Grasshopper will be able to reveal how the teacher intended the question to be answered.
-C2
I think you've faux-pa'd somewhere. The outer loop does not have a constant current so no i4 would be required. There are only 3 distinct links in the circuit (a link being a path from node to node) so there can only possibly be 3 different currents.
Having been taught this about a year and a half ago, I am quite sure that my method is the method required given the question (even if I may have made an error!)
Hopefully Grasshopper will be able to reveal how the teacher intended the question to be answered.
-C2
I think you've faux-pa'd somewhere. The outer loop does not have a constant current so no i4 would be required. There are only 3 distinct links in the circuit (a link being a path from node to node) so there can only possibly be 3 different currents.
Having been taught this about a year and a half ago, I am quite sure that my method is the method required given the question (even if I may have made an error!)
QUOTE (Grasshopper+Oct 20 2008, 09:18 PM)
Interestingly, on the test we had a circuit with actual values for voltage and resistence. I guess he felt sorry for us. Still, I need to get better at this one.
---------------------------------------------------------------------------------------
Sapo, I think your link is dead.
Sorry, sir. It may have been at the time, but not my fault! Did you want my site so you'd be able to make an effective post?
I thought all the clever folks had caught on to this little trick...
My ISP is an effective monopoly, so they don't have much of a desire to minimize response time...
Road Runner: The Power of **** You.
I am still here, and the Joint is slowly growing.
Maybe from Arthritis?
We are happy to have new members, and happier still at the return of prodigals. You know who you might be. You can read between the lines, as clever as you are...
---------------------------------------------------------------------------------------
Sapo, I think your link is dead.
Sorry, sir. It may have been at the time, but not my fault! Did you want my site so you'd be able to make an effective post?
I thought all the clever folks had caught on to this little trick...
My ISP is an effective monopoly, so they don't have much of a desire to minimize response time...
Road Runner: The Power of **** You.
I am still here, and the Joint is slowly growing.
Maybe from Arthritis?
We are happy to have new members, and happier still at the return of prodigals. You know who you might be. You can read between the lines, as clever as you are...
I agree there are three nodes. Grasshopper has headed straight for a current loop analysis which suggests (to me) that this is the intended mode. I could (of course) be wrong.
-C2.
-C2.
QUOTE (Confused2+Oct 22 2008, 09:12 PM)
I agree there are three nodes. Grasshopper has headed straight for a current loop analysis which suggests (to me) that this is the intended mode. I could (of course) be wrong.
-C2.
No, there are 2 nodes. I think you need to re-draw what you think the circuit looks like and re-evaluate it.
-C2.
No, there are 2 nodes. I think you need to re-draw what you think the circuit looks like and re-evaluate it.
I agree .. two nodes. Can you generate the matrix?
Edit .. I would need to convert the (three) given voltage sources into equivalent current sources to attempt what you propose .. perhaps my method is inelegant and you have a better one. I am sure there are many methods.
Edit .. I would need to convert the (three) given voltage sources into equivalent current sources to attempt what you propose .. perhaps my method is inelegant and you have a better one. I am sure there are many methods.
QUOTE (Confused2+Oct 22 2008, 09:23 PM)
I agree .. two nodes. Can you generate the matrix?
Edit .. I would need to convert the (three) given voltage sources into equivalent current sources to attempt what you propose .. perhaps my method is inelegant and you have a better one. I am sure there are many methods.
I think that given my current stance it would be rude for me to not try!
Let me go through my old notes tonight to refresh myself on the method and I will post back.
Edit .. I would need to convert the (three) given voltage sources into equivalent current sources to attempt what you propose .. perhaps my method is inelegant and you have a better one. I am sure there are many methods.
I think that given my current stance it would be rude for me to not try!Let me go through my old notes tonight to refresh myself on the method and I will post back.
Hi all,
I've been away for a while, so this thread has become a bit "stale" by this time, but here's a reply anyway:
There are only 3 currents because there are only 3 branches (also called links), and there are only 2 voltages because there are only 2 nodes.
Using the outer loop as well as the two smaller loops will do no good, for the mathematical reason that the Kirchoff voltage law equation from the outer loop is the sum of the equations from the two smaller loops. Therefore this equation is not linearly independent with the other two, and so it is redundant. Physically, the path around the outer loop is what you get if you start from the bottom node and trace clockwise around the left loop, then backtrack up the middle wire, and continue around the right loop. Backtracking causes the voltage changes along the shared central wire to cancel out.
There are never more independent voltage law equations than the number of minimal size loops needed to cover the circuit. Therefore, you get only two voltage equations, with three variables.
The third equation must therefore come from Kirchoff's current law. There are two nodes, and one node is always redundant (because the sum of all currents out of every node but the last is the same as the sum of the currents into that last node, so the currents at that node sum to zero automatically if they sum to zero at the other nodes). Therefore, you only need to write the current law for one node. At the top node, you have I1 + I2 = I3. Together with the two voltage equations, you have a solvable system of 3 equations in 3 unknowns.
Other methods: The loop current method reduces the variables to 2 by automatically incorporating the current law. Similarly, the node voltage method reduces the variables to 1 by automatically incorporating the voltage law. These methods also have the advantage that there is a nice formula for the matrix elements, which lets you write them down directly into the matrix form, just by looking at the circuit.
I haven't heard of "fluence graphs" by that name, but I'm aware of a method that avoids matrices by constructing a "tree diagram" for each voltage or current source. Perhaps it is similar.
Undoubtedly the course has moved on since you posted this question, but if you still would like to see the automatic matrix method, I'd be glad to post it.
Hope that helps!
--Stuart Anderson
I've been away for a while, so this thread has become a bit "stale" by this time, but here's a reply anyway:
There are only 3 currents because there are only 3 branches (also called links), and there are only 2 voltages because there are only 2 nodes.
Using the outer loop as well as the two smaller loops will do no good, for the mathematical reason that the Kirchoff voltage law equation from the outer loop is the sum of the equations from the two smaller loops. Therefore this equation is not linearly independent with the other two, and so it is redundant. Physically, the path around the outer loop is what you get if you start from the bottom node and trace clockwise around the left loop, then backtrack up the middle wire, and continue around the right loop. Backtracking causes the voltage changes along the shared central wire to cancel out.
There are never more independent voltage law equations than the number of minimal size loops needed to cover the circuit. Therefore, you get only two voltage equations, with three variables.
The third equation must therefore come from Kirchoff's current law. There are two nodes, and one node is always redundant (because the sum of all currents out of every node but the last is the same as the sum of the currents into that last node, so the currents at that node sum to zero automatically if they sum to zero at the other nodes). Therefore, you only need to write the current law for one node. At the top node, you have I1 + I2 = I3. Together with the two voltage equations, you have a solvable system of 3 equations in 3 unknowns.
Other methods: The loop current method reduces the variables to 2 by automatically incorporating the current law. Similarly, the node voltage method reduces the variables to 1 by automatically incorporating the voltage law. These methods also have the advantage that there is a nice formula for the matrix elements, which lets you write them down directly into the matrix form, just by looking at the circuit.
I haven't heard of "fluence graphs" by that name, but I'm aware of a method that avoids matrices by constructing a "tree diagram" for each voltage or current source. Perhaps it is similar.
Undoubtedly the course has moved on since you posted this question, but if you still would like to see the automatic matrix method, I'd be glad to post it.
Hope that helps!
--Stuart Anderson
Thanks mr_homm.
I would very much like to see the matrix solution (and any others you would care to demonstrate), if you don't mind. You are correct, we have since moved on (to differential equations), but on the test I didn't get the entire problem correct, so I could stand to see it.
I would very much like to see the matrix solution (and any others you would care to demonstrate), if you don't mind. You are correct, we have since moved on (to differential equations), but on the test I didn't get the entire problem correct, so I could stand to see it.
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