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Simple potentiometer circuit help

  1. Feb 18, 2006 #1
    Hi ,please Can I have some suggestions to do this exercice:

    A simple potentiometer circuit used to measure unknown voltages accurately is shown below.
    Here Vs is the known source voltage, Vxis the unknown voltage, and the resistor is a variable one from which the values R1 and R2 can be read from the position of the pointer. these resistances are varied until the current in the ammeter is zero. Show that the unknown voltage then has the value Vx=Vs*R2/(R1+R2).


    I can not create a picture.
    Thank you

    B
     
  2. jcsd
  3. Feb 18, 2006 #2
    How can we give you correct suggestion without seeing your picture?
     
  4. Feb 18, 2006 #3

    Gokul43201

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    Gold Member

    Brad, please read the guidelines for posting in the coursework forum.
     
  5. Feb 18, 2006 #4
    [​IMG]

    Sorry, I really tried the problem.
    What I did is to write the equations for the two loops.

    Loop 1:
    -Vs +i1*R1+i2*R2=0

    Loop2:
    Vx-i2*R2=0

    Then I find i2=-Vx/R2

    Then replace i2 in the equation of the loop1.
    My problem is the ammeter. I know that it has an intern resistor ,but introducing it in the equation will cause new unknown to appear in the equation...

    This is why I am asking help.

    Thank you
     
  6. Feb 18, 2006 #5
    Here's my solution to your problem

    [tex]V_S=I(R_1+R_2)[/tex]

    [tex]V_X=IR_2[/tex](because the current through ammeter is [tex]0[/tex]).

    Then the answer is obvious[tex]V_X=\frac{V_SR_2}{R_1+R_2}[/tex]
     
  7. Feb 18, 2006 #6

    Thank you, but please can you give me more precision about Vx=I*R2?
    if there is current in the ammeter, this means that Vx should equal to 0 since U=R*I right???
     
  8. Feb 18, 2006 #7
    Using the first and the second Kirchhoff's law you will easily find out what I mean and remember to use [tex]I_A=0[/tex] in your equations.
     
  9. Feb 18, 2006 #8

    lightgrav

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    Homework Helper

    Brad Sue, your current i2 = 0 , once the resistors have been properly adjusted.
    The I in "IR_2" ... (in post#5) is your i1 ... R1 and R2 have same current
    since none of it goes thru the ammeter.
     
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