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General help understanding emf and p.d

  1. May 18, 2009 #1
    Circuit1.jpg

    Graph to show the variation of the p.d. V across the battery with current I as R is varied.


    Table1.jpg

    i) Draw line of best fit.

    ii) Use it to determine the emf of the battery;

    and the internal resistance r of the battery.


    The gradient [itex] \approx \frac{\Delta V}{\Delta I} = 1.67 \Omega [/itex]

    Just don't know what to do really. Need an example and explanation of the concept.
     
  2. jcsd
  3. May 18, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Let the internal resistance of the battery be r.
    The same current,I, passes through both r and R right?

    So by Kirchoff's 2nd law

    the emf of the battery=sum of the pd's around the loop
    If V is the pd across the load then

    E=V+Ir.

    Since they plotted V against I, rearrange the equation it in the form a straight line y=mx+c

    you calculated the gradient correctly to give a resistance. So rearrange the equation and you will see what that gradient represents and what the intercept represents as well.
     
  4. May 18, 2009 #3
    [itex] V = f(I) [/itex]

    [itex] V = -Ir + E [/itex]


    y-intercept is emf, gradient is -r. Thanks.
     
  5. May 19, 2009 #4
    Hi I had a similar question, which I posted at the the thread 'internal resistance graph'. It was basically why are V are I inversely proportional in this graph, but proportional in the graph of an ohmic conductor?
     
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