P28_017.PDF

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Chapter 28 - 28.17
17. (a) Using Eq. 28-4, we take the derivative of the power P = i 2 R with respect to R and set the result
equal to zero:
( R + r ) 2 = E
dP
dR =
d
dR
E
2 R
R )
( R + r ) 3
2 ( r
−
=0
which clearly has the solution R = r .
(b) When R = r , the power dissipated in the external resistor equals
P max = E
2 R
( R + r ) 2
R = r
= E
2
4 r
.
643283085.001.png

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