p33_060.pdf
(
67 KB
)
Pobierz
Chapter 33 - 33.60
60. (a) Thepowerconsumedbythelightbulbis
P
=
I
2
R/
2. Sowemustlet
P
max
/P
min
=(
I/I
min
)
2
=5,
or
I
I
min
2
=
E
m
/Z
min
E
m
/Z
max
2
=
Z
max
Z
min
2
=
R
2
+(
ωL
max
)
2
R
2
=5
.
Wesolvefor
L
max
:
ω
=
2(120V)
2
/
1000W
L
max
=
2
R
2
π
(60
.
0Hz)
=7
.
64
×
10
−
2
H
.
(b) Nowwemustlet
R
max
+
R
bulb
R
bulb
2
=5
,
or
R
max
=(
√
5
1)
R
bulb
=(
√
5
1)
(120V)
2
1000W
=17
.
8Ω
.
This is not done because the resistors would consume, rather than temporarily store, electromag-
neticenergy.
−
−
Plik z chomika:
kf.mtsw
Inne pliki z tego folderu:
p33_002.pdf
(59 KB)
p33_011.pdf
(64 KB)
p33_005.pdf
(66 KB)
p33_018.pdf
(73 KB)
p33_017.pdf
(71 KB)
Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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naruszono regulamin