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E.m.f and internal resistance

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data
    This problem relates to an experiment to determine the e.m.f and internal resistance of a battery using a variable load. Taking values for V & I a graphy of V vs. I was plotted.

    2. Relevant equations

    3. The attempt at a solution
    After plotting the graph and calculating the gradient. I am not sure where the gradient and the equation y=mx+c fit in the equation E=IR+Ir, although I believe it could be rearraging the equation to look like IR=E-Ir. Hence y=IR, E=mx and -Ir=c
  2. jcsd
  3. Oct 22, 2007 #2


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    From the equations yu've given you could have: [itex]E=V+Ir[/itex]. What would the gradient and intercept be then?
  4. Oct 23, 2007 #3


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    Homework Helper

    If you wqrite your original equation in this form

    E = Ir + V

    and rearrange you get

    V = E - rI

    this now corresponds with the linear equation for your graph:

    V = c + mI

    where c is the y-intercept and m is the gradient of your graph.
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