Problem 2

A battery is modeled by a constant emf

E

in series with internal resistance

r>0

and supplies a purely resistive load

R>0

; neglect lead resistance and temperature changes. Let

I

be the circuit current,

U

the terminal voltage (also the load voltage), and

P_R

the load power. (a) Find

I

and

U

in terms of

E,r,R

. (b) Calculate

P_R

and the power dissipated inside the source. (c) Determine the value of

R

that maximizes

P_R

, the maximum power, and the efficiency at that operating point.

problems.proof.stepOf

P_{\max}=\frac{E^2}{4r},\quad \eta=\frac{P_R}{EI}=\frac12

problems.proof.analysis

At the maximum, $R=r$ , so $I=E/(2r)$ and $P_R=E^2/(4r)$ . Efficiency is load power divided by source power $EI$ ; equal resistors make it one half.