I am supposed to take a hard look at this network, come up with a matrix and use it to get all the currents. Can someone help me to setup the equations for the matrix?

- anonymous

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- schrodinger

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- anonymous

This is what I came up with. But according to Wolfram Alpha it can not be solved so I guess my equations to begin with a wrong.

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- anonymous

Can you rotate the picture please? My neck kinda hurts otherwise :P

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- anonymous

o.O
Seriously? Every picture is oriented in the right way.

- anonymous

Oh, I'm new sorry. That's a stupid interface. OK I downloaded it...
Do you know what A and B are supposed to stand for? voltage sources I guess?

- anonymous

Between A and B we got 10 Volts.

- eSpeX

If you set it up using mesh analysis you find that you have two equations and two unknowns.
Start with the left mesh and you get:
\[R_1i_1 + R_5(i_1-i_2) + R_2i_1 = 0\]
and for the right:
\[R_4i_2 + R_5(i_2-i_1) + R_3i_2 = 0\]
Build your matrix from that.

- anonymous

:O
I thought I have 5-6 unknows and as much equations.
You did not consider the currents that run into a knot.

- anonymous

Why don´t you got a i4 and i3?

- eSpeX

Because there are only two meshes (loops).

- eSpeX

Maybe this will help: http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

- anonymous

Yes but that does not mean that the current that runs through r2 is also the current that runs through r4.

- eSpeX

Well no, because a mesh is voltage, r*i, so you know the voltage on each of them and thus you can calculate the current through each element.

- anonymous

But we also do not know the voltage on r2 and r4 so we can not say if there runs the same current or not.
Why don´t you consider also looking at the knots?

- eSpeX

Knots?

- anonymous

A place where the current that goes in equals the current that goes out. The other important part of kirchoffs-rule. Perhaps node is the right word?

- anonymous

The thick dots on the drawing where one line splits into two lines.

- eSpeX

Ah, nodes. The nodes are considered. But you know that the voltage drops around any loop equal zero. If you want to do node analysis you would need to know a little more information like the voltage on each side of your circuit or maybe one of the drops within it.

- anonymous

I know the voltage between A and B is 10 Volts.

- anonymous

And through Kirchhoff I know that the currents in a node add up to zero. I used three nodes and two mashes to get five equations. You can see them in the top.

- eSpeX

Well you only have two nodes where two or more components meet so that would only yield one equation. Kirchoff's current law says that the current entering a node must equal the current leaving the node, so they do not equal zero.

- anonymous

There are four nodes.

- eSpeX

I meant to say, three or more components meet.

- eSpeX

However to use node voltage method, you need a reference node as well.

- anonymous

Well know that all the current that goes in has to come out.

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- anonymous

What do you mean with reference node?

- eSpeX

One that equals zero. Although if you took your 'A' side and called it 0, your 'B' as +10, you could do a node equation for the top, and bot nodes. That would give you four equations and four unknowns I believe.

- eSpeX

Including the mesh equations.

- anonymous

But if you sum up all the currents equal zero for any node. I don´t need a reference node. Otherwise it would mean that there are electrons pilling up in the node if the sum of all the currents entering and exiting the node does not equal zero.

- eSpeX

Perhaps the analysis you are tasked with is using a method that I am unfamiliar with. I am not sure who would be a good person to ask though.

- anonymous

I think I just have to use all of Kirchhoffsrules.

- eSpeX

You are going to have 7 unknowns using KCL, with the information you are given it seems to me that using mesh and node analysis would be the easiest way to approach this.

- anonymous

Why do you get 7 unknowns? I get max. 6. We can conclude that I that comes from A into the network is the same I that goes out of the network into B.

- eSpeX

Accounting for all of the currents I get 7, if you wish to discount your ib=ia then you would have 6, yes.

- anonymous

Did you solve the system? I threw mine into Wolfram Alpha and got weird stuff, for example i1=i2. That seems very weird to me. Also i4 and i5 are booth negative.

- anonymous

Can you perhaps help me to calculate the total resistance of this network?

- phi

Here is how Prof. Strang would do the problem
http://ocw.mit.edu/courses/mathematics/18-085-computational-science-and-engineering-i-fall-2008/video-lectures/lecture-12-graphs-and-networks/
I thought I would apply his approach to your problem.

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