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TomLikesPhysics

  • 2 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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  1. TomLikesPhysics
    • 2 years ago
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    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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  2. BluFoot
    • 2 years ago
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    Can you rotate the picture please? My neck kinda hurts otherwise :P

  3. TomLikesPhysics
    • 2 years ago
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    o.O Seriously? Every picture is oriented in the right way.

  4. BluFoot
    • 2 years ago
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    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?

  5. TomLikesPhysics
    • 2 years ago
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    Between A and B we got 10 Volts.

  6. eSpeX
    • 2 years ago
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    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.

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

  8. TomLikesPhysics
    • 2 years ago
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    Why don´t you got a i4 and i3?

  9. eSpeX
    • 2 years ago
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    Because there are only two meshes (loops).

  10. eSpeX
    • 2 years ago
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    Maybe this will help: http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

  11. TomLikesPhysics
    • 2 years ago
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    Yes but that does not mean that the current that runs through r2 is also the current that runs through r4.

  12. eSpeX
    • 2 years ago
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    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.

  13. TomLikesPhysics
    • 2 years ago
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    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?

  14. eSpeX
    • 2 years ago
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    Knots?

  15. TomLikesPhysics
    • 2 years ago
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    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?

  16. TomLikesPhysics
    • 2 years ago
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    The thick dots on the drawing where one line splits into two lines.

  17. eSpeX
    • 2 years ago
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    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.

  18. TomLikesPhysics
    • 2 years ago
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    I know the voltage between A and B is 10 Volts.

  19. TomLikesPhysics
    • 2 years ago
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    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.

  20. eSpeX
    • 2 years ago
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    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.

  21. TomLikesPhysics
    • 2 years ago
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    There are four nodes.

  22. eSpeX
    • 2 years ago
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    I meant to say, three or more components meet.

  23. eSpeX
    • 2 years ago
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    However to use node voltage method, you need a reference node as well.

  24. TomLikesPhysics
    • 2 years ago
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    Well know that all the current that goes in has to come out.

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  25. TomLikesPhysics
    • 2 years ago
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    What do you mean with reference node?

  26. eSpeX
    • 2 years ago
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    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.

  27. eSpeX
    • 2 years ago
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    Including the mesh equations.

  28. TomLikesPhysics
    • 2 years ago
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    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.

  29. eSpeX
    • 2 years ago
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    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.

  30. TomLikesPhysics
    • 2 years ago
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    I think I just have to use all of Kirchhoffsrules.

  31. eSpeX
    • 2 years ago
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    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.

  32. TomLikesPhysics
    • 2 years ago
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    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.

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

  34. TomLikesPhysics
    • 2 years ago
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    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.

  35. TomLikesPhysics
    • 2 years ago
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    Can you perhaps help me to calculate the total resistance of this network?

  36. phi
    • 2 years ago
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    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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