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Calculating current in an unbalanced wheatstone bridge?

  1. Oct 25, 2015 #1
    Capture_zpsxfss8vvz.png I am having troubles understanding how to calculate the current at RSensor when RSensor is 110 Ohms. Any help would be appreciated.
  2. jcsd
  3. Oct 25, 2015 #2


    Staff: Mentor

    Here is your procedure. ((100+100) in parallel with (110+100))+270+270 is the total resistance. Can you do the rest?
  4. Oct 25, 2015 #3
    In your circuit the current is being fed from two terminals kept at say +V and -V volts so effectively you are driving the circuit by a potential difference of say 2V volts - the you apply the Ohms law and calculate the current in the two arms of a loop which does not have a source of EMF-so you can apply Kirchhoff's loop equation and can get a relation between the different current in the two arms. as the R-sensor differs by 10 ohms only from the other arm the difference in current will come out to be pretty small.
    well i do not guess where lies the problem of calculation-
    you have to use one relation P.D. (2V) = sum of Current X Resistances if you are traversing a path from one terminal to another;
    and the other relation is in a looop sum of currentX resistances = the EMF of source =0
    The two arms of the loop gives you equivalent resistance 1/Req = 1/200 + 1/ 210'
    if you move on those lines the current in Rsensor comes to approx. 2/55 amp. and in other arm about 21/550 amp.
    well i do not know whetheri could interpret your problem correctly!
  5. Oct 29, 2015 #4
    Unbalanced Wheatstone Bridge.jpg
    In my opinion, the actual Wheatstone Bridge circuitry is as in attached sketch.

    In this case, the unknown 5 parameters will be I1,I2,Ix[through Rsensor],I4,Ig[through Ammeter].

    The 5 equations are:





  6. Oct 29, 2015 #5
    Neglecting Rg[Rg=0] Ix=0.008895 A
  7. Oct 30, 2015 #6

    The Electrician

    User Avatar
    Gold Member

    Using the loop method, only 3 equations are needed. Choosing the loop currents I1, I2 and I3 like this:


    we then set up the 3 equations in matrix form like this, including the resistance of the ammeter Rg to obtain a more general solution:


    I2 (green) is the same as Ix, and I3 (blue) is the current through the ammeter. Setting Rg = 0, we obtain Ix = 8.895 mA.

    If the ammeter is missing (Rg = ∞), then Ix = 9.112 mA.
    Last edited: Oct 30, 2015
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