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Variable Resistror, Calculating Vo

  1. Mar 15, 2012 #1
    Screen shot 2012-03-15 at 6.24.50 PM.png 1. The problem statement, all variables and given/known data
    A variable resistor, R, is connected to the terminals of a battery (dashed box in the diagram). The graph below shows data collected in the circuit as R is varied. What is V0, in volts?
    What is the internal resistance, r, in ohms?

    2. Relevant equations

    Ohm's Law

    3. The attempt at a solution

    I know that for this case, Vo will be found where the current is calculated to be at its lowest.
    from the graph that i am given, It is not clear where it will equal zero, as for the internal resistance, i know that I=ε/r
    so
    -r=dV/dI
    the slope of a V vs I graph will give the internal resistance.
     

    Attached Files:

    Last edited: Mar 15, 2012
  2. jcsd
  3. Mar 15, 2012 #2

    gneill

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    Staff: Mentor

    Do you have a circuit diagram that you can post? Your graph shows current versus resistance; is the resistance that of the variable resistor? How were they measured or determined?
     
  4. Mar 15, 2012 #3
    oops! i forgot to click attach on the second file.
    new here :)
     
  5. Mar 15, 2012 #4

    gneill

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    Staff: Mentor

    No problem :smile: I see a voltmeter in the circuit, but no current meter. How were the values for the current versus resistance plot obtained?
     
  6. Mar 15, 2012 #5
    these were given to me! this is all the information given.
     
  7. Mar 15, 2012 #6

    gneill

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    Okay, fair enough. In that case, I'd suggest that the first thing to do would be to write an expression for the current in the circuit assuming that the cell has voltage Vo, with resistors R and r as shown.
     
  8. Mar 15, 2012 #7
    okay, so i have the equation I=V/R

    since the resistors are in parallel,
    (1/Req)=(1/R+1/r)
    take the inverse of both sides to get Req
    so i get
    I=Req(Vo)

    Is this right?
     
  9. Mar 15, 2012 #8

    gneill

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    The resistors are not in parallel; the battery cell is between two of their leads so they are not connected to each other at both ends.
     
  10. Mar 15, 2012 #9
    okay, so it would still be
    I=Req(Vo)
    but instead i would just add the resistances without inversing them.
    Req= R+r
    right?
     
  11. Mar 15, 2012 #10

    gneill

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    Hmm. Not quite. Ohm's law: V = I*R, so that I = V/R.
    Sure. But write the equation without replacing R and r with Req; you want to have "access" to those variables as you proceed.
     
  12. Mar 15, 2012 #11
    okay yes, ( i saw my bad re arranging in the equation)

    So I will end up with
    I=Vo/(R+r)

    I am still in the mist here... i am not sure how to calculate each individual resistor since i am not given their individual values.
     
  13. Mar 15, 2012 #12

    gneill

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    Ha! :smile: That's where the graph comes in. You can read off convenient pairs for R and I and build yourself a couple of equations using the expression above as a template. Two unknowns require two equations.
     
  14. Mar 16, 2012 #13
    !!!!!!!!!!
    thank z, me and my friend used I=Vo/(R+r) to solve for Vo,
    Vo=I(R+r)
    got two equations
    Vo=5.25(r) *since when R=0, I=5.25
    Then i used another point
    Vo=1.5(2.5+r)
    and solved for Vo.
    With this, i was able to solve for my internal resistance to equal one.
    :) thank you gneill!
     
  15. Mar 16, 2012 #14

    gneill

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    Staff: Mentor

    Glad to help! :smile:
     
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