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Help with Wheatstone Bridge Strain Gauge Circuit

  1. Mar 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Given your average Wheatstone Bridge with R1 R2 and R3 unknown resistor values and R4=RG= strain gauge, how to solve for the resistor values given only input voltage and the value of the resistances between the resistors?



    2. Relevant equations
    Not Sure



    3. The attempt at a solution
    Tried writing differential circuit equation, was told I was way off base. Answer is simple, apparently. Help!
     
  2. jcsd
  3. Mar 6, 2015 #2

    gneill

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    Staff: Mentor

    Between which resistors? And what are the locations of the resistors R1, R2, R3 in the bridge? A diagram would be helpful.
     
  4. Mar 6, 2015 #3
    Each resistor is separated by the same value, ~ 750 Ohms.
    http://web.deu.edu.tr/mechatronics/TUR/strain_gauge.htm [Broken]
    This is a general picture, but fits the circuit to a T.
     
    Last edited by a moderator: May 7, 2017
  5. Mar 6, 2015 #4

    gneill

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    Staff: Mentor

    Okay, much better.

    You've used terms like "between" and "separated by" when referring to resistance values. That description isn't forming a clear picture for me. Can you give an example of where you might place an Ohmmeter's leads in order to take such a measurement? Do you mean if meter was placed across any one of the resistors in the bridge (assuming that the voltage source has been disconnected for the measurement) that it would read 750 Ohms?
     
  6. Mar 6, 2015 #5
    We clamped at the end of one resistor and the beginning of the other. Imagine the resistances encompassing the points of the bridge.
     
  7. Mar 6, 2015 #6

    gneill

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    Your descriptions are still very loose and open to interpretation. I have no way to tell which end of a resistor is the "end" or the "beginning". Since resistors are tied together, any node could serve as both ends and beginnings for two different resistors. So. Would it be fair to say that this is an example of what you mean:

    Fig1.gif
     
    Last edited: Mar 6, 2015
  8. Mar 7, 2015 #7
    Yes, that is almost exactly what is meant, except that the resistances were taken across the spaces between resistors rather than across them. Imagine moving the red lead to the nearest end of R1 and the black lead to the other end of R2, a slight counterclockwise shift for both leads. Then there is no resistor between them, but one on each side. Sorry to be vague.
     
  9. Mar 7, 2015 #8

    gneill

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    Staff: Mentor

    Like this then:

    Fig2.gif

    Presumably you must open the circuit at node a? Otherwise you've got a short circuit across the meter leads due to the wiring between R1 and R2, and the meter would read zero Ohms.
     
  10. Mar 7, 2015 #9
    Yes, I think that's how it works!
     
  11. Mar 7, 2015 #10

    gneill

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    Staff: Mentor

    Okay, so the meter is just reading the sum of all the resistances. Each reading is giving you the same result, so no new information is obtained by making more than one of these readings.

    It would have been different if you had measured across each resistor, since then you would have serial and parallel combinations to work with.

    I think the best you can do with just the one data point is to assume that the bridge is designed to be balanced when strain gauge is not being flexed, and that the potential at the nodes where the voltmeter attaches is half the supply voltage. What would that tell you about the resistors?
     
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