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Tricky V I Graph Need help please

  1. Jun 6, 2013 #1
  2. jcsd
  3. Jun 6, 2013 #2

    rude man

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    What do you mean by 'gradient'?

    You might consider taking more than 1 measurement of I and V ...
     
    Last edited: Jun 7, 2013
  4. Jun 6, 2013 #3

    berkeman

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    As the load resistor value gets smaller, the loop current increases, and the measured source voltage decreases because of the increasing voltage drop across the voltage source's internal resistance.

    Does that make sense? Can you now show us your work on the graph?
     
  5. Jun 7, 2013 #4
    I drew a line that made a triangle from the V axis to the I axis, so when the current was at its highest V was 0 and when V was at its highest I was 0, so it was a negative slope.

    I used emf= V+Ir and rearranged it go get emf - Ir = V. Im confused at wether the graph should be the a positive slope since the voltmeter is measuring the battery and not the load resistor if you get where im coming from
     
  6. Jun 7, 2013 #5
    So im saying I drew it like this http://www.s-cool.co.uk/a-level/assets/learn_its/alevel/physics/Resistance/internal-resistance-emf-and-potential-difference/Finding%20the%20internal%20resistance.gif [Broken]

    Does it make a difference to the graph that were measuring V at a different point
     
    Last edited by a moderator: May 6, 2017
  7. Jun 7, 2013 #6

    rude man

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    Your graph is fine as far as it goes.

    I don't understand your question abot V being measured "at a different point". The problem shows exactly where V is measured - across the battery including its internal resistance. Your eqation V = emf - I*r fully corresponds to the diagram of the problem.

    Now, how do you propose to construct your graph? You don't know emf or r. You're supposed to use the graph to determine emf and r.
     
    Last edited by a moderator: May 6, 2017
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