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Wheatstone Bridge with Thevenin's Theorem

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img685/602/homeworkprobsg29.jpg [Broken]


    The bridge circuit provides a good illustration of the use of Thevenin's theorem. Find the current through the bridge resistor (50Ω).


    2. Relevant equations

    V = IR

    source transformations,

    current division, voltage division, KCL, KVL


    3. The attempt at a solution

    Some trouble understanding this whole procedure, but I think you cut out the 100V so it becomes short circuited and then cut out the bridge resistor (50Ω) out of the picture and then the wire with the bridge resistor becomes open-circuited? (With the bridge resistor acting as a "load resistor"?)


    Then the rest of the bridge resistors, after adding in series, become 1000Ω and 3000Ω in parallel.

    Which is then R = 750Ω equivalent.

    I think it is also supposed to be the Thevenin resistance too

    So RTh = 750Ω

    then I think you have 750Ω and the original 50Ω resistor back in series, I'm stuck with what to substitute back as the original source (unless I already did something else wrong too). I think I'm supposed to use current division / voltage division later on to find the current through the 50Ω.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 19, 2013 #2

    gneill

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    Staff: Mentor

    Yes, you're looking to find the Thevenin equivalent so you want to find the potential across the open terminals (where you've removed the 50Ω load resistor) and find the equivalent resistance of the network looking into those terminals.

    You'll want to be careful determining what's in series and what's in parallel here once you've shorted (suppressed) the voltage source; Explain your reasoning.

    Hint: To find the potential across the open terminals, first find the individual potentials at those nodes.
     
  4. Feb 19, 2013 #3
    So I got this:

    http://imageshack.us/a/img72/8300/homeworkprobsg29edit.jpg [Broken]


    Haven't forgotten everything about wheatstone bridges.

    and it looks like in the middle of each branch I think it's (50V - 50V) on the left and then

    (66.66V - 33.33V) on the right side

    (If that's how you do it)

    then wouldn't the V across the terminals be (33V - 0V) ?

    But where do I go from here? Is R(Thevenin) still 750Ω ? What would happen now?
     
    Last edited by a moderator: May 6, 2017
  5. Feb 19, 2013 #4

    gneill

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    Staff: Mentor

    Close, but no. Pick a common reference node and watch out for the polarities of the potential drops. The bottom rail (connected to the battery negative terminal) would be a good choice for reference node. What are the potentials at the terminals with respect to that common node? What then is the difference between those potentials?
    Rth isn't 750Ω. Look again at the topology of the circuit when the battery is suppressed (replaced by a short circuit).
     
    Last edited by a moderator: May 6, 2017
  6. Feb 19, 2013 #5
    I don't see what you mean... if the bridge resistor is taken out, the left and right branches go in parallel, so the drops on each branch should still add to 100V though, right?

    Unless somehow it's something like:

    50V - (-50V) and 33V - (-66V), which I can't see / distinguish, even if there was a reference node on the bottom.

    And how would I get R(Thevenin) then too? Isn't the bridge resistor taken out?
     
  7. Feb 19, 2013 #6

    gneill

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    Staff: Mentor

    What will the meters read?
    attachment.php?attachmentid=55923&stc=1&d=1361324784.gif
    The bridge resistor is taken out. The top and bottom of the diamond shaped circuit are shorted together by the suppressed voltage source. If they are shorted together, they comprise one node. You can move all the connections to that node to one location...
     

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  8. Feb 19, 2013 #7
    I'm guessing then 50 on the left and either 33 or 66 on the right, depending on which way you're going then. (66V still if going from top to bottom?)

    I thought the Thevenin resistance would just be the simplified resistors with the 50V taken out and the bridge wire becoming open-circuited.


    And after I find it, what would I do afterwards?
     
    Last edited: Feb 19, 2013
  9. Feb 19, 2013 #8

    gneill

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    Staff: Mentor

    Don't guess. Look at where the leads of a meter connect. What is the potential that lies between the leads? You've marked it on the components that parallel the meter leads! Think of taking a "KVL walk" from the reference node to the node in question; that gives you the potential of that node.
    The bridging resistor is removed and the voltage source suppressed. You want the resistance of the resulting network as "seen" from the the open terminals where the bridging resistor was removed. I think you understand that well enough. However, so far you haven't properly grasped the topology of the circuit to identify what's in parallel and what's in series.

    Remember that you can (graphically) move the connections of components to any location on a wire that comprises a node that they connect to. In this case the suppressed voltage source joins the top and bottom nodes INTO A SINGLE NODE so you can move all the connections to it anywhere you wish along that continuous wire path. So move the connection points of the resistors at the bottom point to the top (or vice versa if you wish). Now how does the circuit appear?
    Once you have the Thevenin voltage and Thevenin resistance, draw the Thevenin equivalent and stick the load resistor onto it. Find the current.
     
  10. Feb 19, 2013 #9
    This might sound dumb but if I were to do a KVL walk would I have to include the 100V source?

    So with that reference node would it be 50V for V1 and then 66V for V2 on the figure there.

    And that voltage across the bridge (Thevenin voltage) would be V1 - V2 ?
     
  11. Feb 19, 2013 #10

    gneill

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    You could if you wish. Any path that gets you from the negative lead to the positive lead of the meter will do. However, there's a very short path available, and you know the potential across the component...
    Yes to the first, no to the second. Write out your KVL walk. Include all the the potential changes along the path (you've already penciled in the various potential changes for all the components). There's a very short path...
    Yes it would.
     
  12. Feb 19, 2013 #11
    So it would be 100V - 50V for the first and then 100V - 33V for the second one

    making v2 = 66V then, right?
     
  13. Feb 19, 2013 #12

    gneill

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    Staff: Mentor

    That'll do it. Or you could simply have "walked" across the resistors between the leads of the meters: 50V for V1 and 66.6V for V2.
     
  14. Feb 19, 2013 #13
    So :

    V(Th) = -16.67V


    Then R(Th) would be with respect to one end of the bridge to other so

    500 and 500 are in parallel and 1000 and 2000 are in parallel

    so 250 + 666.67 = 916.667

    R(Th) = 916.667 ohm

    R(load) = 50 ohm


    So assuming V(Th), R(Th), and R(load) are all in series then I do voltage division:

    V across 50 ohm =[ 50 / (50 + 916.667) ]*(-16.67V)

    = -0.862V

    then I = V / R

    so I = -0.862/50

    I across the bridge circuit = -0.0172A

    and it's the right answer.

    Could it have been V2 - V1 instead though for the voltage across the bridge? The answer would then have an opposite sign and that's what I did first.
     
  15. Feb 19, 2013 #14

    gneill

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    Good!
    Since Rth and RL are in series, you could have simply summed them and applied Ohm's Law to the total. The same current flows through both.
    It depends only upon what the problem statement says; if they want a particular current it's up to them to tell you which direction it's to be assumed to flow. If they want a particular potential difference they need to specify what direction (polarity) is to be assumed for positive values.
     
  16. Feb 20, 2013 #15
    Thanks
     
    Last edited: Feb 20, 2013
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