Question

If a resistance of R ohms is connected across a battery of E volts with internal resistance (the resistance within the battery) r ohms, then the power in watts dissipated in the external resistor is given by P = (E^2R)/((R+r)^2. If E and r are fixed, find R so that the power dissipated externally is a maximum.

Answer 1

@Zarkon @mathstudent55 @Hero @iambatman

Answer 2

It is an optimization problem for calculus...

Answer 3

Have you studied implicit differentiation? This looks like it would be generally unpleasant to explicitly solve for R, then to differentiate.
I would clear the fraction, implicitly differentiate the function, then set dR equal to zero. This should give you a relatively simple function of E and r.

Answer 4

Or you could do a Google search of "Maximum Power Transfer Theorem"

Answer 5

The power in the external resistance R is given by:
\[\large P=E^{2} \times \frac{R}{(R+r)^{2}}\]
\[\large \frac{dP}{dR}=E^{2} \times \frac{(R+r)^{2}-R.2(R+r)}{(R+r)^{4}}\]
For a maximum or a minimum value of P, dP/dR = 0

Answer 6

This happens when the numerator of the right hand side is zero, which occurs when:
\[\large (R+r)^{2}=2R(R+r)\]
This is satisfied by R = r.
In order to demonstrate that this actually represents a maximum value for P, you must differentiate again and note the sign of d^2P/dR^2. Are you able to do this final step?

Answer 7

I misread the question. kropot covered it well, though I thought we were supposed to guide to an answer, not solve the problem.

Answer 8

@ZippySlug Thank you. Please note that I have left it for @ComputerNerd to finish by differentiating again to demonstrate that the tentative solution represents a maximum value for P.

Answer 9

An excellent response kropot72, I thought a diagram and another idea may be helpful.
Expanding dividing numerator and denominator by RL gettig:
It is clear that The load resistor RL would equal the Internal resistance Rint for maximum transfer of power. Note that the efficiency is only 50% as you are only getting 1/2 the power generated into the load. If you are looking for maximum efficiency, matching RL to Ri isn't the way to go.