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

problems.proof.prevStep problems.proof.nextStep

U=IR=E-Ir=\frac{ER}{R+r}

problems.proof.analysis

The load is connected directly across the source terminals, so $U=IR$ . Subtracting the internal drop $Ir$ from $E$ gives the same value and checks the sign.

problems.proof.pitfall. Constraint condition: Do not confuse terminal voltage

U

with emf

E

under load: the internal resistance causes a drop

Ir