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Wheatstone bridge

  1. May 25, 2017 #1
    1. The problem statement, all variables and given/known data
    Set voltage between 1. a-b; 2- b-c; 3 c-d 4. a-c; if voltage of battery is 12V and R_1=2000 ohms R_2=3000 ohms; R_3= 4000 ohms. (see attached file)

    2. Relevant equations
    voltage=final voltage- innitial voltage
    resistors in series R_1,2=R_1+R_2
    parallel resistors: R_1,2=(R_1*R_2)/(R_1+R_2)

    3. The attempt at a solution
    I have no idea how to solve such problem. I can use Kirchhoff laws in circuits, I can use all equations I wrote in "relevant equations", I watched video how to set equivalent resistance of those resistors through "delta to ypsilon transformation" () but I have absolutely no idea how to apply my knowledge in such problem.
     

    Attached Files:

    Last edited by a moderator: May 25, 2017
  2. jcsd
  3. May 25, 2017 #2

    gneill

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    Staff: Mentor

    Good. So that means you can write the KVL loop equations for the circuit (or the KCL node equations). What do you get?
     
  4. May 25, 2017 #3
    Well, resistance of equivalent resistor would be R_eq=2461.44 ohms.
    KVL loop looks: +voltage of battery-(current*equivalent resistor)=0 => 12volts-current*2461.44ohms=0 => current=0.004875ampers
    So what should I do next?
     
  5. May 25, 2017 #4

    gneill

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    Staff: Mentor

    You have not solved the circuit. You need to either solve for loop currents (using KVL loop equations) or for node voltages (KCL node equations). My own preference would be to write node equations for the two essential nodes (b and c) since the problem is asking for potential difference between nodes.
     
  6. May 25, 2017 #5
    You mean:
    -I_3*R_1-I_2*R_3+I_1*R_2=0
    -I_2*R_3-I_4*R_1+I_5*R_2=0
    I_1+I_2=I_4
    I_3=I_2+I_5
    ?
    upload_2017-5-26_1-38-10.png
     
  7. May 25, 2017 #6

    gneill

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    Staff: Mentor

    That's a start. You need one more loop, one involving the voltage source.
     
  8. May 26, 2017 #7
    There should be two loops involving source:
    ε-R_1*I_3-R_2*I_5=0
    ε-R_2*I_1-R_1*I_4=0
     
  9. May 26, 2017 #8

    gneill

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    Staff: Mentor

    Okay, but you'll only need one of them; you need just enough loops so that every component is included in at least one of the loops. After that any additional loops will not introduce any new (independent) information to the system of equations.
     
  10. May 26, 2017 #9
    Ok, so now I choose 5 equations with 5 unknown currents, find them and using Ohm´s law I will find voltage on resistors?
     
  11. May 26, 2017 #10

    gneill

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    Staff: Mentor

    Yes. Three loop equations and two node equations (KCL at two of the nodes).

    If I may make a suggestion, you can reduce the number of current unknowns if you are careful about how you introduce new currents when you label your circuit. Proceed systematically and only introduce the minimum number of new currents required at junctions. For example, if one current (say ##i_1##) enters a junction and two leave, then only introduce one new current (say ##i_2##) leaving the junction by one of the paths so that ##i_1 - i_2## leaves by the other. If you do it this way you're effectively doing the KCL sums ahead of time and you'll end up with the minimum number of current variables required and you'll only require loop currents to solve.

    upload_2017-5-26_6-49-59.png
     
  12. May 26, 2017 #11
    OK, thanks a lot.
     
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