thwing (jrt2329) – Homework 05 – Yao – (59110)
1
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printout
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have
22
questions.
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before answering.
001
10.0 points
The current in a wire decreases with time
according to the relationship
I
= (1
.
54 mA)
×
e
−
at
where
a
= 0
.
13328 s
−
1
.
Determine
the
total charge
that
passes
through the wire from
t
= 0 to the time the
current has diminished to zero.
Correct answer: 0
.
0115546 C.
Explanation:
I
=
dq
dt
q
=
integraldisplay
t
t
=0
I dt
=
integraldisplay
∞
t
=0
(0
.
00154 A)
e
−
0
.
13328 s
−
1
t
dt
= (0
.
00154 A )
e
−
0
.
13328 s
−
1
t

0
.
13328 s
−
1
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
∞
0
=
0
.
0115546 C
.
002
10.0 points
If 5
×
10
21
electrons pass through a 20 Ω resis
tor in 14 min, what is the potential difference
across the resistor?
Correct answer: 19
.
0714 V.
Explanation:
Let :
n
= 5
×
10
21
,
Δ
t
= 14 min
,
R
= 20 Ω
,
and
q
e
= 1
.
602
×
10
−
19
C
.
The current is defined as
I
=
Δ
Q
Δ
t
.
Since each electron carries a
q
e
, we have
I
=
n q
e
Δ
t
=
(
5
×
10
21
)
(1
.
602
×
10
−
19
C)
14 min
parenleftbigg
1 min
60 s
parenrightbigg
= 0
.
953571 A
.
Then from Ohm’s Law,
V
=
I R
= (0
.
953571 A) (20 Ω) =
19
.
0714 V
.
003
10.0 points
A conductor with crosssectional area 4 cm
2
carries a current of 9 A.
If the concentration of free electrons in the
conductor is 8
×
10
28
electrons
/
m
3
, what is
the drift velocity of the electrons?
Correct answer: 0
.
00175542 mm
/
s.
Explanation:
Let :
I
= 9 A
,
n
= 8
×
10
28
electrons
/
m
3
,
q
e
= 1
.
60218
×
10
−
19
C
,
and
A
= 4 cm
2
= 0
.
0004 m
2
.
The current in a conductor is given by
I
=
n q v
d
A ,
where
n
is the number of charge carriers per
unit volume,
q
is the charge per carrier,
v
d
is
the drift velocity of the carriers and
A
is the
cross section of the conduction.
Solving for
v
d
, we have
v
d
=
I
n q A
=
9 A
8
×
10
28
electrons
/
m
3
×
1
1
.
60218
×
10
−
19
C
×
1
0
.
0004 m
2
×
1000 mm
1 m
=
0
.
00175542 mm
/
s
.
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thwing (jrt2329) – Homework 05 – Yao – (59110)
2
004
(part 1 of 3) 10.0 points
Consider two cylindrical conductors made of
the same ohmic material. Conductor 1 has a
radius
r
1
and length
ℓ
1
while conductor 2 has
a radius
r
2
and length
ℓ
2
.
Denote:
The currents of the two conductors
as
I
1
and
I
2
, the potential differences between
the two ends of the conductors as
V
1
and
V
2
,
and the electric fields within the conductors
as
E
1
and
E
2
.
V
1
vector
E
1
I
1
ℓ
1
r
1
V
2
vector
E
2
I
2
ℓ
2
r
2
If
ρ
2
=
ρ
1
, r
2
= 2
r
1
, ℓ
2
= 3
ℓ
1
and
V
2
=
V
1
, find the ratio
R
2
R
1
of the resistances.
1.
R
2
R
1
=
1
4
2.
R
2
R
1
= 3
3.
R
2
R
1
=
2
3
4.
R
2
R
1
= 2
5.
R
2
R
1
= 4
6.
R
2
R
1
=
3
4
correct
7.
R
2
R
1
=
1
2
8.
R
2
R
1
=
4
3
9.
R
2
R
1
=
3
2
10.
R
2
R
1
=
1
3
Explanation:
The relation between resistance and resis
tivity is given by
R
=
ρ ℓ
A
=
ρ ℓ
π r
2
.
Then since
r
2
= 2
r
1
and
ℓ
2
= 3
ℓ
1
, the ratio
of the resistances is
R
2
R
1
=
ρ ℓ
2
π r
2
2
π r
2
1
ρ ℓ
1
=
ℓ
2
r
2
1
ℓ
1
r
2
2
=
(3
ℓ
1
)
r
2
1
ℓ
1
(2
r
1
)
2
=
3
4
.
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 Spring '09
 YAO
 Physics, Current, Work, Resistor, Correct Answer, Electrical resistance, Series and parallel circuits

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