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20
.
2
ir
iR
i
rR
ε
−− = ⇒ =
+
The power dissipated in
R
is
Pi
R
R
==
+
()
.
2
4
2
2
2
2
We find the maximum by setting the derivative with respect to
R
equal to zero. The
derivative is
dP
dR
r
R
R
=
+
−
+
=
−
+
4
2
16
2
42
2
2
3
2
3
2
3
εε
.
The derivative vanishes (and
P
is a maximum) if
R = r
/2. With
r
= 0.300
Ω
, we have
0.150
R
=Ω
.
(b) We substitute
R = r
/2 into
P
= 4
ε
2
R
/(
r
+ 2
R
)
2
to obtain
22
2
max
2
4
( / 2)
(12.0 V)
240 W.
[
2( / 2)]
2
2(0.300 )
r
P
rr
r
=
=
+Ω
41. (a) The batteries are identical and, because they are connected in parallel, the
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 Spring '08
 Any
 Physics, Current

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