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

\boxed{I=\frac{E}{R+r},\ U=\frac{ER}{R+r},\ P_R=\frac{E^2R}{(R+r)^2};\ R_{\max}=r,\ P_{\max}=\frac{E^2}{4r},\eta_{\max}=\frac12}

problems.proof.analysis

Current, voltage, and power have dimensions A, V, and W. The load power tends to zero for either very large or very small resistance, confirming the interior maximum.