Update, I think I figured this one out, please let me know if I made a mistake:
b) R_L = 0
E = Qemf
c) Took the derivative of the power and set it equal to 0 to get R_L = r. This would mean it gets half of the energy produced by the battery.
d) In order to get 9/10 of the energy produced by the...
Ah I see, that makes it more straight forward. Given I can only get smaller as R_L increases then wouldn't the largest value of I arise from R_L having a value of 0? That appears trivial though. Also, I'm assuming the energy would simply be Q*emf?
Problem 5: Current, Energy and Power A battery of emf ε has internal resistance i R ,
and let us suppose that it can provide the emf to a total charge Q before it expires.
Suppose that it is connected by wires with negligible resistance to an external (load) with
resistance L R .
a) What is...