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Can you rotate the picture please? My neck kinda hurts otherwise :P
o.O Seriously? Every picture is oriented in the right way.
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?
Between A and B we got 10 Volts.
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.
:O I thought I have 5-6 unknows and as much equations. You did not consider the currents that run into a knot.
Why don´t you got a i4 and i3?
Because there are only two meshes (loops).
Maybe this will help: http://www.allaboutcircuits.com/vol_1/chpt_10/3.html
Yes but that does not mean that the current that runs through r2 is also the current that runs through r4.
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.
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?
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?
The thick dots on the drawing where one line splits into two lines.
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.
I know the voltage between A and B is 10 Volts.
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.
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.
There are four nodes.
I meant to say, three or more components meet.
However to use node voltage method, you need a reference node as well.
What do you mean with reference node?
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.
Including the mesh equations.
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.
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.
I think I just have to use all of Kirchhoffsrules.
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.
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.
Accounting for all of the currents I get 7, if you wish to discount your ib=ia then you would have 6, yes.
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.
Can you perhaps help me to calculate the total resistance of this network?
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.