P28_029.PDF

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Chapter 28 - 28.29
29. (a) The batteries are identical and, because they are connected in parallel, the potential differences
across them are the same. This means the currents in them are the same. Let i be the current in
either battery and take it to be positive to the left. According to the junction rule the current in
R is 2 i and it is positive to the right. The loop rule applied to either loop containing a battery and
R yields
i = E
E−
ir
−
2 iR =0 =
⇒
r +2 R .
The power dissipated in R is
P =(2 i ) 2 R =
2 R
( r +2 R ) 2
4
E
.
We find the maximum by setting the derivative with respect to R equal to zero. The derivative is
dP
dR =
2
( r +2 R ) 2 −
4
E
2 R
( r +2 R ) 3
16
E
=
4
2 R )
( r +2 R ) 3
2 ( r
−
.
The derivative vanishes (and P is a maximum) if R = r/ 2.
(b) We substitute R = r/ 2into P =4
E
2 R/ ( r +2 R ) 2 to obtain
P max =
2 ( r/ 2)
[ r +2( r/ 2)] 2
4
E
= E
2
2 r
.
E
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