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Multiloop circuit: kirchhoff

  • Thread starter gills
  • Start date
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1. Homework Statement
YF-26-61.jpg


What is the current in I(1), I(2), and I(3)?

2. Homework Equations

At nodes [tex]\sum[/tex]I = 0

Loops [tex]\sum[/tex]V = 0



3. The Attempt at a Solution

Ok, I'm definitely new to this multiloop stuff so i'm not quite sure what the best way to work these problems is.

Here's what i have for nodes:

I(4) + I(6) - I(5) = 0
I(2) - I(4) - I(3) = 0
I(1) + I(3) - I(6) = 0


and loops:

12 - I(4)5 + I(6)8 - 9 + I(1)1 - I(2)1 = 0
12 - I(3)10 - 9 + I(1)1 - I(2)1 = 0
9 - I(6)8 - I(1) = 0
12 - I(4)5 - I(2)1 = 0


Any help on the best and easiest way to work these would be great. Thanks
 

Answers and Replies

360
21
Are there labels missing on the diagram. I(4), I(5), I(6)
 
116
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Are there labels missing on the diagram. I(4), I(5), I(6)
they aren't on the diagram. I just ended up putting them in myself because it seems like i would need them. However, i probably overcomplicated it as i always tend to do. Do i even need an I(4), I(5), and I(6)?
 
454
0
IMO the most efficient way to solve these is the following method.

1. replace series and parallel resistances (doesn't work in this one)

2. give an arbitrary point a potential of 0. I'd take the node in the middle here between
the four resistances. If you need an equivalent resistance, call the 2 connections to the circuit 0 and V.

3. guess the potential at all points where more than 2 wires meet and call them V_1, V_2 etc. There are two such points here

4. use ohms law to work out all currents as a function of these potentials.

5. Use kirchhof's current rule in each of the points from step 3. This will give you as
many equations as unknowns.

6. Now that you found the potentials, compute the currents you need.

The loop rule is completely redundant here.
 
Doc Al
Mentor
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1,084
Though kamerling gives an excellent approach, there's nothing wrong with your original method using nodes and loops. For this problem I would use three independent loop equations (because I see three loops) and two node equations.

Since your I(5) just runs through a piece of bare wire, I would not have defined it. (You can always shrink a bare wire down to a point.) Thus I would only have 5 currents and thus only need 5 equations.
Here's what i have for nodes:

I(4) + I(6) - I(5) = 0
I(2) - I(4) - I(3) = 0
I(1) + I(3) - I(6) = 0
I would just drop your first equation. (Forget I(5).)

and loops:

12 - I(4)5 + I(6)8 - 9 + I(1)1 - I(2)1 = 0
12 - I(3)10 - 9 + I(1)1 - I(2)1 = 0
9 - I(6)8 - I(1) = 0
12 - I(4)5 - I(2)1 = 0
You really only have three independent loops. Your first equation is just the sum of the last two (or it should be--you have some errors).

So I would drop your first equation. The last two need to be corrected: You have the sign in front of the I(1) and I(2) terms wrong.

Once you have your corrected 5 equations, it's just a matter of turning the crank. 5 equations and 5 unknowns. Start by eliminating one variable--I(4) say--in all your equations. Then eliminate another--I(6) say. And so on until you've solved for one of the currents. Then go back and find the others.
 
116
0
Though kamerling gives an excellent approach, there's nothing wrong with your original method using nodes and loops. For this problem I would use three independent loop equations (because I see three loops) and two node equations.

Since your I(5) just runs through a piece of bare wire, I would not have defined it. (You can always shrink a bare wire down to a point.) Thus I would only have 5 currents and thus only need 5 equations.

I would just drop your first equation. (Forget I(5).)


You really only have three independent loops. Your first equation is just the sum of the last two (or it should be--you have some errors).

So I would drop your first equation. The last two need to be corrected: You have the sign in front of the I(1) and I(2) terms wrong.

Once you have your corrected 5 equations, it's just a matter of turning the crank. 5 equations and 5 unknowns. Start by eliminating one variable--I(4) say--in all your equations. Then eliminate another--I(6) say. And so on until you've solved for one of the currents. Then go back and find the others.

Ok, i've rewritten my equations and my diagram because i've made the wire in between a node where there's no component like you said.

nodes:
I(2) - I(3) - I(5) = 0
I(1) + I(3) - I(4) = 0

loops:
12 - 5I(5) - I(2) = 0
9 - 8I(4) - I(1) = 0
12 - 10I(3) - 9 + I(1) - I(2) = 0

5 eq, 5 unknowns. Unfortunately (and fortunately ;)) i've plugged these into MATlab to check and see if it would be right and it was:

I(1) = 0.848
I(2) = 2.14
I(3) = 0.171

However, i need to try and solve these on my own. What is usually the most effective way for systems of equations, substitution or elimination? Just such a tedious task....

Thanks Doc and everyone else
 
Doc Al
Mentor
44,828
1,084
While I'm sure there are clever ways of solving systems of equations (using matrix methods or whatever), I suggest using the simple method of Gaussian elimination. Just systematically eliminate the variables until you have one left. Keep all your intermediate equations, since once you get one value you'll be able to quickly get the others just by plugging in.

Tedius? You bet! Get busy. :smile:
 
116
0
While I'm sure there are clever ways of solving systems of equations (using matrix methods or whatever), I suggest using the simple method of Gaussian elimination. Just systematically eliminate the variables until you have one left. Keep all your intermediate equations, since once you get one value you'll be able to quickly get the others just by plugging in.

Tedius? You bet! Get busy. :smile:
thanks Doc. Tedious indeed, but i got it.
 

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