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Internal Resistance of a Real Battery + Resistance of the Circuit

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data
    (a) In the figure shown, show that the rate at which energy is dissipated in R as thermal energy is a maximum when R = r. (b) Show that this maximum power is P = EMF2/4r.

    http://www.practicalphysics.org/imageLibrary/jpeg400/208.jpg [Broken]

    The only difference between this picture and the one in my book is that they specify the direction of current going one way (clockwise) throughout the whole circuit.



    2. Relevant equations

    P = i2r

    EMF - ir - iR = 0

    P = EMF2/4r



    3. The attempt at a solution

    (b) switching EMF - ir - iR = 0 so that it is equal to i:

    i = EMF / (r + R)

    and assuming r = R:

    i = EMF / 2r

    substituting this in for i in P = i2r, you get:

    P = [ EMF / 2r ]2r
    P = EMF2r/4r2
    P = EMF2/4r

    (a) for r = R, as shown above:

    P = EMF2/4r

    and for r =/= R:

    P = EMF2r/(r2 + 2rR + R2)

    Thing is, I don't really understand how the rate of dissipation of thermal energy is greatest at r = R. Isn't this just a linear relationship? Wouldn't the power be greatest if the internal resistance was zero, as in an ideal battery?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 18, 2009 #2
    See "maximum power theorem" at wikipedia.

    Max power would be delivered to the load resistor if internal resistance were zero, but the question is about maximum heat dissipated.
     
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