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17.
(a) Using Eq. 284, we take the derivative of the power
P
=
i
2
R
with respect to
R
and set the result
equal to zero:
dP
dR
=
d
dR
µ
E
2
R
(
R
+
r
)
2
¶
=
E
2
(
r
−
R
)
(
R
+
r
)
3
=0
which clearly has the solution
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Unformatted text preview: 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 ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Power

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