Internal Resistance of a Real Battery + Resistance of the Circuit

  • Thread starter ohgeecsea
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  • #1
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Homework Statement


(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.



Homework Equations



P = i2r

EMF - ir - iR = 0

P = EMF2/4r



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?
 
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Answers and Replies

  • #2
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See "maximum power theorem" at wikipedia.

Wouldn't the power be greatest if the internal resistance was zero, as in an ideal battery?
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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