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anonymous
 3 years ago
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
 3 years ago
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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you rotate the picture please? My neck kinda hurts otherwise :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0o.O Seriously? Every picture is oriented in the right way.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Between A and B we got 10 Volts.

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2If 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: \[R_1i_1 + R_5(i_1i_2) + R_2i_1 = 0\] and for the right: \[R_4i_2 + R_5(i_2i_1) + R_3i_2 = 0\] Build your matrix from that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0:O I thought I have 56 unknows and as much equations. You did not consider the currents that run into a knot.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why don´t you got a i4 and i3?

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2Because there are only two meshes (loops).

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2Maybe this will help: http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes but that does not mean that the current that runs through r2 is also the current that runs through r4.

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2Well 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
 3 years ago
Best ResponseYou've already chosen the best response.0But 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A place where the current that goes in equals the current that goes out. The other important part of kirchoffsrule. Perhaps node is the right word?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The thick dots on the drawing where one line splits into two lines.

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2Ah, 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
 3 years ago
Best ResponseYou've already chosen the best response.0I know the voltage between A and B is 10 Volts.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And 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
 3 years ago
Best ResponseYou've already chosen the best response.2Well 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
 3 years ago
Best ResponseYou've already chosen the best response.0There are four nodes.

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2I meant to say, three or more components meet.

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2However to use node voltage method, you need a reference node as well.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well know that all the current that goes in has to come out.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What do you mean with reference node?

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2One 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
 3 years ago
Best ResponseYou've already chosen the best response.2Including the mesh equations.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But 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
 3 years ago
Best ResponseYou've already chosen the best response.2Perhaps 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
 3 years ago
Best ResponseYou've already chosen the best response.0I think I just have to use all of Kirchhoffsrules.

eSpeX
 3 years ago
Best ResponseYou've already chosen the best response.2You 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
 3 years ago
Best ResponseYou've already chosen the best response.0Why 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
 3 years ago
Best ResponseYou've already chosen the best response.2Accounting for all of the currents I get 7, if you wish to discount your ib=ia then you would have 6, yes.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did 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
 3 years ago
Best ResponseYou've already chosen the best response.0Can you perhaps help me to calculate the total resistance of this network?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0Here is how Prof. Strang would do the problem http://ocw.mit.edu/courses/mathematics/18085computationalscienceandengineeringifall2008/videolectures/lecture12graphsandnetworks/ I thought I would apply his approach to your problem.
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