# Internal Resistance of a Real Battery + Resistance of the Circuit

## 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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