Question

If a resistor of R ohms is connected across a battery of E colts with internal resistance r ohms, then the power (in watts) in the external resistor is \[P = \frac{E^2R}{(R+r)^2}\]
If E and r are fixed but R varies, what is the maximum value of the power?

Answer 1

Take the first derivative and use the quotient rule.

Answer 2

hmm..

Answer 3

so \[P' = \frac{2(R+r)E^2R - (R+r)^2E^2}{(R + r)^4}\]
???

Answer 4

Lets rearrange this: E^2R^2-E^2r^2/(R+r)^2 is simplified. When we expand -(R+r)^2E^2 it is -E^2(R^2 + 2rR +r^2). The first term expands to 2E^2R^2+2E^2rR. Notice R term drops out. Lets go back to the top and you have a simple difference of the squares which will allow you to find the zeros and hence critical points like max mins. I hope this helps. We do not have to worry about the denominator.

Answer 5

yes indeed.