Power in Electric Circuits — Question 6

Question 6

Maximum load power and maximum efficiency are different objectives

A source with voltage parameter $\mathcal E$ and internal resistance r drives load R. The supplied current relation is $I=\mathcal E/(R+r)$. Internal loss is $I^2r$.

1. Find the R that maximizes load power.

2. Find efficiency at that R and the smallest R that gives efficiency at least 90 percent.

Question 6 - Solution

Maximum load power and maximum efficiency are different objectives

(a) Method and calculation

$$P_L=\frac{\mathcal E^2R}{(R+r)^2},\quad\frac{dP_L}{dR}=\frac{\mathcal E^2(r-R)}{(R+r)^3}.$$$$R_{\max P}=r,\quad P_{\max}=\mathcal E^2/(4r).$$

(b) Interpretation and further calculation

$$\eta=\frac{I^2R}{I^2(R+r)}=\frac R{R+r},\quad\eta(R=r)=1/2.$$$$R/(R+r)\ge.9\Rightarrow R\ge9r.$$

Physical check

Efficiency approaches one at large R while load power approaches zero; the objectives do not coincide.