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Chapter 28
149
P28.53
I
Rr
=
+
ε
, so
P
==
+
IR
R
2
2
2
af
or
R
+=
F
H
G
I
K
J
2
2
.
Let
x
≡
2
, then
x
R
a
f
2
or
x
R
r
22
20
+−
−
=
.
With
r
=
120
.
Ω
, this becomes
Rx
R
2
240
144 0
−
=
..
a
f
,
which has solutions of
R
xx
=
−−
±
−
−
576
2
2
.
a
f
a
f
.
(a)
With
=
920
V
and
=
12 8
. W,
x
=
661
.
:
R
=
+±
−
=
421
2
384
2
...
.
Ω
or
0375
Ω
.
(b)
For
=
V
and
=
21 2
. W
,
x
≡=
2
399
.
R
=
−
=
±−
159
2
322
2
2
a
f
.
The equation for the load resistance yields a complex number, so
there is no resistance
that will extract 21.2 W from this battery. The maximum power output occurs when
Ω
, and that maximum is:
max
.
2
4
17 6
r
W.
P28.54
Using Kirchhoff’s loop rule for the closed loop,
+
−−=
12 0 2 00
4 00
0
.
II
, so
I
=
200
A
VV
ba
−=
+
−
−
=
−
4 00
2 00
4 00
0 10 0
4 00
.
V
A
V
a
f
ΩΩ
.
Thus,
∆
V
ab
=
400
.
V and
point
is at the higher potential
a
.
P28.55
(a)
RR
eq
=
3
I
R
=
3
series
I
R
2
3
(b)
R
RRR
R
eq
=
++
=
1
111
3
bg
I
R
=
3
parallel
I
R
3
2
(c)
Nine times more power is converted in the
parallel connection.
*P28.56
(a)
We model the generator as a constantvoltage power supply.
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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