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A Kirchoff's Law problem with a bridge circuit

  1. Sep 12, 2014 #1
    1. The problem statement, all variables and given/known data

    I want to use Kirchoff's current law to analyze the pictured curcuit and find the current drawn fro the voltage source between the positive terminal and node 1.

    [PLAIN]http://s42.photobucket.com/user/Jemspak/media/circuitdiagram.jpg.html" [Broken]

    (i can't figure out how to get the image insert to work)

    2. Relevant equations

    I a using V=IR and I know that the current into any node has to be equal to the current out, so the algebraic sum is zero. I also know that at the other end of the circuit the current that re-enters the power source has to be the same as that which left it.

    For our purposes [itex]V_0[/itex] is 24 V. Al the resistors are 10Ω.

    3. The attempt at a solution

    So here are the equations I am using. I am trying to find out if I set up the problem correctly, because I feel like I am getting an extra unknown.

    At node 1:

    [itex]I_{source}-\frac{V_0-V_2}{R_2}-\frac{V_0-V_1}{R_1}=0[/itex]

    node 2:
    [itex]\frac{V_0-V_2}{R_2}-\frac{V_2-V_4}{R_4}-\frac{V_2-V_3}{R_3}=0[/itex]

    node 3:
    [itex]\frac{V_0-V_1}{R_1}-\frac{V_1-V_3}{R_3}-\frac{V_3-V_6}{R_6}=0[/itex]

    node 4:

    [itex]\frac{V_2-V_4}{R_4}-\frac{V_4-V_5}{R_5}=0[/itex]

    node 5:
    [itex]\frac{V_1-V_6}{R_6}-\frac{V_4-V_5}{R_5}+I_{source}=0[/itex]

    And assuming I set this up correctly I should get the following system of equations:

    [itex]10I_{source}-V_1-V_2=48[/itex]
    [itex]-3V_2-V_4-V_3=-24[/itex]
    [itex]-3V_1-V_3-V_6=-24[/itex]
    [itex]-V_2-V_5=0[/itex]
    [itex]-V_1+V_4-V_5-V_6+10I_{source}=0[/itex]

    I know that [itex]V_1+V_2+48 = 10I_{source}[/itex] so the last equation becomes

    [itex]V_4-V_5-V_6+V_2=-48[/itex]

    ANd the system now looks like this
    [itex]10I_{source}-V_1-V_2=48[/itex]
    [itex]-3V_2-V_4-V_3=-24[/itex]
    [itex]-3V_1-V_3-V_6=-24[/itex]
    [itex]-V_2-V_5-=0[/itex]
    [itex]V_4-V_5-V_6+V_2=-48[/itex]

    I notice that V5=-V2 and I can sub that into the last equation as well getting me V4-V6+2V2=-48.

    Well and good, but i suspect I set this up wrong and that is why my system of equations is looking awfully complicated. So any help in seeing where I err is much appreciated.

    this might end up double posting because I was trying to get the image to show.
     

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    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 12, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Do you have two threads with the same question?? Message a moderator and ask for one thread to be deleted.
     
  4. Sep 12, 2014 #3
    yeah I couldn't get the image thing to work at first and I thought i lost the thing the first time...
     
  5. Sep 12, 2014 #4

    gneill

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    Staff: Mentor

    Which thread do you wish to keep? I'll remove the redundant one.
     
  6. Sep 12, 2014 #5
    keep this one, it has a better picture (I think)
     
  7. Sep 12, 2014 #6

    gneill

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    Done.

    Regarding your question, note that your node labelled (4) is a non-essential node --- The two resistors R4 and R5 can be combined into a single resistance. Also note that node (1) has a fixed potential with respect to the reference node thanks to the 24V DC source. So, if you can find the potentials at nodes 2 and 3, you will be in a position to find the currents through R1 and R2.

    One more thing, in nodal analysis you don't need to write a node equation for the reference node. It just becomes a redundant equation. So really you only have two essential nodes, nodes 2 and 3, which you need to write equations for :wink:
     
  8. Sep 12, 2014 #7
    Thnks a lot. Just curious if I at least had the equations at all right-- I noticed myself that one of the equations had to be redundant (I couldn't remember if it was OK to simply add the resistances in R4 and R5). And I did wonder that the two equations with I in them seemed to be exactly equal and if I could ignore them as a result... but then I seem to end up with 3 equations and 4 unknowns, though that may be mitigated by combining the R4 and R5 piece into a single circuit element of current (V2-V5)/(V4+V5). Yes?
     
  9. Sep 12, 2014 #8
    By the way sine node 1 potential is fixed would that mean the current is: I(source) - V0/R2 - V0/R1 = 0?
     
  10. Sep 12, 2014 #9

    NascentOxygen

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    Where the figure has a voltage and current associated with each resistor, I assume that voltage to be the voltage across that resistor. Most methods of analysis use the node voltages, so you should relabel your figure with node voltages to suit your method.

    You didn't set it up correctly at the start. You went with the labelling given, but mistook the resistor voltages to be node voltages and ended up in a proper pickle. You have some node voltages going by 2 names.

    Label node potentials with one name and keep to it.

    .... transferred from duplicate thread.
     
  11. Sep 12, 2014 #10
    So if you could help me out (and this might be my conceptual problem) wha is the difference between node voltage and that across the resistor? I thought whatever current is coming in to the node will be V/R, and that voltage wil be whatever is coming out of the previous resistor. Said voltage will be V(incoming to Resistor) - V(coming out). That's wrong, I guess, so what is the proper way to do t?
     
  12. Sep 12, 2014 #11

    gneill

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    Your equations have some problems. In nodal analysis you want to deal with the node potentials only, and they are all taken to be potentials with respect to the reference node. So you don't want to introduce extraneous variables like V6. The potential across R6 with respect to the reference node is V3 - 0, the difference between the potentials at the terminals of the resistor (the reference node is defined to be at 0 Volts).

    So the current associated with the R6 branch at node 3 is simply V3/R6.

    You might find it convenient to write your node equations as a sum of currents all flowing into the node or all flowing out (choose one or the other). This way you don't have to bother with choosing one current to be flowing in and the others flowing out. Myself, I almost always choose to write the equations assuming that all currents are assumed to be flowing out of the given node. The math will automatically take care of the signs.

    Nope. The bottom ends of R1 and R2 are not at ground potential. They have potentials V3 and V2. The potential difference across R1 is thus 24 - V3, and across R2 it's 24 - V2.
     
  13. Sep 12, 2014 #12
    OK, I just want ot check though, if i assume all current is flowing into the node what about the rule (KCL) that says they have to add up to zero? I still have to pick one as negative, no?
     
  14. Sep 12, 2014 #13

    gneill

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    Nope. The math will take care of that for you. Just write your branch currents using the potential difference due to the node voltages at the ends of the branch. So, for example, suppose I were writing the equation for node 2 in your circuit. I choose to assume that all currents are flowing out of node 2. That implies that I am making the assumption that node 2 has a higher potential than the nodes around it. So the term for the current flowing out of node 2 through R3 to node 3 would be (v2 - v3)/R3.

    That potential difference, v2 - v3, enforces the assumption of the current direction. Even if the assumption doesn't match the reality of the actual current direction, the math will sort it out for you via the v2 being greater than v3 or v3 turning out to be greater than v2 when you solve for them.
     
  15. Sep 12, 2014 #14

    gneill

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    Voltage doesn't "come in" and "come out" of a component. Current does that.

    Voltage is a difference in potential between two points. In the case of the voltage across a component, it's the potential difference between the two terminals of that component. So think of it as being "across" rather than "though".

    For a node potential, it's the potential at that node with respect to the reference node. Again note that it's across two points, one being the node itself and the other being the reference node.


    attachment.php?attachmentid=73080&stc=1&d=1410539244.gif
     

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    Last edited: Sep 12, 2014
  16. Sep 13, 2014 #15
    OK looking at this let me know if I got it right this time for my system of equations:

    [itex]I_{source} - \frac{V_s - V_3}{R_1}-\frac{V_s - V_2}{R_2}=0[/itex] - current at node 1
    [itex]\frac{V_3-V_2}{R_3} - \frac{V_s - V_3}{R_1}-\frac{V_s - V_2}{R_2}=0[/itex] - current at node 2
    [itex]\frac{V_3-V_2}{R_3} - \frac{V_s - V_3}{R_1}-\frac{V_3}{R_6}=0[/itex] - current at node 3 and we are taking voltage at the last node to be zero

    [itex]\frac{V_3}{R_6} - \frac{V_2}{R_4+R_5}-I_{source} =0[/itex] because the currents have to be the same going back into the power source as coming out.

    Is this system correct? That would mean I can just plug in the values and I should get something sensible. I have been working on this a bit and I am getting answers anyway that seem closer to what my circuit simulator says they ought to be, but I want to be sure.
     
  17. Sep 13, 2014 #16

    gneill

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    The second term in the above equation appears to belong to node 3 rather than node 2.

    The rest seems to be fine.

     
  18. Sep 13, 2014 #17
    Yay. that means I am getting there! Thanks a lot/ FOr some reason this seemed a hard way to do this, using the loop analysis seemed much easier.

    EDIT: Your diagram with the voltmeters for some reason made it clearer.
     
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