Problem 2

An ideal emf source

\mathcal E

in series with internal resistance

r

supplies a resistive load

R

; all components and wires remain at fixed temperature and the circuit is in steady state. (a) Find the current, source terminal voltage, and load power. (b) Find the value of

R

for maximum load power and that maximum. (c) At the maximum-power condition, compare power delivered by the source with power dissipated internally and state the load-power efficiency.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

P_{\rm source}=\mathcal EI=\frac{\mathcal E^2}{2r},\quad P_r=I^2r=P_R,\quad \eta=\frac{P_R}{P_{\rm source}}=\frac12

problems.proof.analysis

For equal resistances carrying the same current, load power equals internal loss. Half the source power reaches the load, so efficiency is $1/2$ .

problems.proof.pitfall. Maximum load power does not mean maximum efficiency. At

R=r

, half the source power is dissipated internally.