segovia (cs39966) – Homework 6 – Spurlock – (14441)
1
This
printout
should
have
21
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
How much energy is stored in a 73 mH induc
tor at the instant when the current is 3
.
2 A?
Correct answer: 0
.
37376 J.
Explanation:
Let :
L
= 73 mH = 0
.
073 H
and
I
= 3
.
2 A
.
The energy is
U
=
1
2
L I
2
=
1
2
(0
.
073 H)(3
.
2 A)
2
=
0
.
37376 J
.
002
10.0points
An inductor of 240 turns has a radius of 7 cm
and a length of 29 cm.
The
permeability
of
free
space
is
1
.
25664
×
10

6
N
/
A
2
.
Find the energy stored in it when the cur
rent is 0
.
6 A.
Correct answer: 0
.
000691596 J.
Explanation:
Let :
r
= 7 cm = 0
.
07 m
,
ℓ
= 29 cm = 0
.
29 m
,
N
= 240 turns
,
I
= 0
.
6 A
,
and
μ
0
= 1
.
25664
×
10

6
N
/
A
2
.
The inductance is
L
=
μ
0
N
2
A
ℓ
=
μ
0
N
2
π r
2
ℓ
= (1
.
25664
×
10

6
N
/
A
2
)
×
(240 turns)
2
π
(0
.
07 m)
2
0
.
29 m
= 0
.
0038422 H
.
The energy stored in an inductor is
U
=
1
2
L I
2
=
1
2
(0
.
0038422 H)(0
.
6 A)
2
=
0
.
000691596 J
.
003
10.0points
A “slinky” toy spring has a radius of 3
.
4 cm
and an inductance of 330
μ
H when extended
to a length of 0
.
6 m.
How many turns are in the spring?
Correct answer: 208
.
293 turns.
Explanation:
Let :
r
= 3
.
4 cm = 0
.
034 m
,
L
= 330
μ
H = 0
.
00033 H
,
and
ℓ
= 0
.
6 m
.
From
L
=
μ
0
N
2
A
ℓ
, we find
N
=
radicalBigg
L ℓ
μ
0
A
=
radicalBigg
(0
.
00033 H)(0
.
6 m)
(1
.
25664
×
10

6
n
/
A
2
)
π
(0
.
034 m)
2
=
208
.
293 turns
.
004(part1of3)10.0points
An inductor and a resistor are connected
with a double pole switch to a battery as
shown in the figure.
The switch has been in position
b
for a long
period of time.
200 mH
6
.
13 Ω
7
.
6 V
S
b
a
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segovia (cs39966) – Homework 6 – Spurlock – (14441)
2
If the switch is thrown from position
b
to position
a
(connecting the battery), how
much time elapses before the current reaches
155 mA?
Correct answer: 4
.
35739 ms.
Explanation:
Let :
R
= 6
.
13 Ω
,
L
= 200 mH
,
and
E
= 7
.
6 V
.
L
R
E
S
b
a
The time constant of an
RL
circuit is
τ
=
L
R
=
0
.
2 H
6
.
13 Ω
= 0
.
0326264 s
.
The final current reached in the circuit is
I
0
=
E
R
=
7
.
6 V
6
.
13 Ω
= 1
.
2398 A
.
The switch is in position
a
in an
RL
circuit
connected to a battery at
t
= 0 when
I
= 0.
Then the current vs. time is
I
=
I
0
parenleftBig
1

e

t / τ
parenrightBig
.
Solving the above expression for
t
, when
I
=
I
1
gives
t
1
=

τ
ln
parenleftbigg
1

I
1
I
0
parenrightbigg
=

(0
.
0326264 s) ln
parenleftbigg
1

0
.
155 A
1
.
2398 A
parenrightbigg
=
4
.
35739 ms
.
005(part2of3)10.0points
What is the maximum current in the inductor
a long time after the switch is in position
a
?
Correct answer: 1
.
2398 A.
Explanation:
After a long time compared to
τ
, we have a
d.c.
circuit with a battery supplying an
emf
E
,
which is equal to the voltage drop
I R
across
the resistor. Thus
I
=
E
R
=
7
.
6 V
6
.
13 Ω
=
1
.
2398 A
.
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 Fall '11
 HOFFMAN
 Inductance, Energy, Work, Inductor, RL circuit, Segovia

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