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Wheatstone Bridge

  1. Sep 16, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    nGMdP.jpg

    Find the detector current i_d in the unbalanced bridge in the figure if the voltage drop across the detector is negligible. The source voltage v_g is 180V.

    2. Relevant equations



    3. The attempt at a solution

    I am not sure what to do, or even how to find the equivalent resistance. I have learned the delta-y transformation, but I don't suspect that will help me here, since the detector has negligible resistance, right? (and if it doesn't have negligible resistance, than the current is negligible, if i am understanding V=IR correctly)

    Someone to point me in the right direction would be greatly appreciated.
     
  2. jcsd
  3. Sep 16, 2012 #2

    rude man

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    If thevoltage drop across the detector is zero, are you entitled to run a wire across it?
     
  4. Sep 16, 2012 #3

    ElijahRockers

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    no but the question is asking for the current there. or at least that's how i interpret the question... i am somewhat new to this but the diagram clearly has a wire going across and it seems to be asking for the current there.

    if there was no wire it would be a simple matter of series/parallel
     
  5. Sep 16, 2012 #4

    rude man

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    It's not a wire, it's a detector, but you are still right. You can think of it as a wire.

    In which case you have only 1 dependent node, so now you can determine the currents thru all the resistors. Then write KCL equations (Ʃi = 0) at both ends of the wire to get the wire current.
     
  6. Sep 18, 2012 #5

    ElijahRockers

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    Ok, well I must have missed something crucial. I have finished all the homework except for this question, using node voltage and mesh current analysis. This question looks comparatively simple, but it is simply not computing for me.....

    I assumed the wire was a node, and solved for the equivalent resistance. I got Req=(175/11)kΩ which makes the current running through the voltage source 11.314mA.

    Then I used node-voltage analysis. V1 is the node connected to the positive terminal of the voltage source, the 6k, and the 30k resistors. V2 is the node connected to the negative terminal of the voltage source, the 12k and the 20k resistors. The reference node is the node in the middle.

    Doing KCL for V1 (11.314mA going in, i1 and i2 going out), I got:
    11.314E-3 = (V1/6k){this is i1} + (V1/30k){this is i2}

    Doing KCL for V2 (all currents are flowing out of this node), I got:
    11.314E-3 + (V2/12k){this is i4} + (V2/20k){this is i3} = 0

    V1 = 56.57V
    V2 = -84.855V

    Using these, I could find each current:

    i1 = 9.428mA
    i2 = 1.886mA
    i3 = -4.243mA
    i4 = -7.071mA

    Now doing a KCL on the reference tells me that between i1 and i3 there is 2.357mA going through the detector, but the software is telling me that this is incorrect. I only have more attempt so I want to be sure that I have the answer before I submit it.

    Node-voltage and mesh current have made things much easier to understand for me, but this question is really giving me hell, and everyone else seems to think it's the easiest question..

    Thanks for your help in advance.
     
  7. Sep 18, 2012 #6

    rude man

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    Well, 1st thing you know is that V1 - V2 must = 180V & yours adds up to only 141.425V.

    You did not get the total resistance across the voltage source right. The total current supplied by the voltage source is more than 11.314 mA.

    But you have the right idea, computing i1 thru i4. Just do that over.
     
  8. Sep 18, 2012 #7

    ElijahRockers

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    EDIT: Wow ok, idk how I ended up with 175/11 k for Req, but I just did the calculations again and came up with 12.5k.

    SUPER EDIT: I see what I did. Instead of 12*20/(12+20), I did 12*20/(12+10). Wasn't taking my time...
     
    Last edited: Sep 18, 2012
  9. Sep 18, 2012 #8

    rude man

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    Tell me what you got for i1 thru i4 and I'll tell you if it's what I got.
     
  10. Sep 18, 2012 #9

    rude man

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    Yah, that's better!
     
  11. Sep 18, 2012 #10

    ElijahRockers

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    Ok.. I redid all the work with the new values and got the correct answer on my last attempt. Thank you very much for noticing that error. I can't believe that was the issue, I'm usually very careful, I promise!
     
  12. Sep 18, 2012 #11

    gneill

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    Three suggestions for approaching the problem:

    1. Assume the detector is a wire and use mesh current analysis to find the current in the wire (there are three loops).

    2. Assume the detector has some resistance R and solve for the current expression using nodal analysis. Then take the limit as R goes to zero.

    3. Treat the left and right sides of the bridge as separate voltage dividers driven by separate voltage sources Vg. Find Thevenin equivalents for both. Connect them output to output and determine the current that flows.
     
  13. Sep 18, 2012 #12

    rude man

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    Nobody's perfect in this most perfect of all worlds!
     
  14. Sep 18, 2012 #13

    ElijahRockers

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    The first two methods you mentioned, node-voltage and mesh current I just learned in class earlier today. A lot of the homework involved Delta-Y/∏-T simplifications which we learned last week.

    We haven't learned about Thevenin equivalents and I am really so busy with modern physics and engineering lab reports that I don't have the time to read ahead, but with this HW out of the way, I can focus on linear algebra and physics (due tomorrow), study for tomorrows physics exam, then hopefully I will have some space to write the lab reports due on thurs and friday, then maybe I can read ahead.. I have never felt so rushed, this semester is getting crazy! Wish me luck

    Thanks for all of your guys' help!
     
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