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60. (a) The power consumed by the light bulb is
P = I
2
R
/2. So we must let
P
max
/
P
min
=
(
I
/
I
min
)
2
= 5, or
I
I
Z
Z
Z
Z
RL
R
m
m
min
max
/
/
.
F
H
G
I
K
J
=
F
H
G
I
K
J
=
F
H
G
I
K
J
=
+
F
H
G
G
I
K
J
J
=
22
2
2
2
2
5
ε
ω
b
g
We solve for
L
max
:
L
R
/
.
.
==
=
×
−
2
2 120
1000
26
0
0
764 10
2
2
VW
Hz
H.
bg
b
g
π
(b) Yes, one could use a variable resistor.
(c) Now we must let
RR
R
,
+
F
H
G
I
K
J
=
bulb
bulb
2
5
or
..
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Power, Light

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