Version One – Homework 5 – Savrasov – 39821 – Apr 16, 2007
This print-out should have 8 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. The due time is Central
time.
2 s to t2 = 4 s?
Correct answer: 33.3333 C.
Explanation:
Let : a = 2 C/s3 ,
b = 3 C/s2 ,
c = 7 C/s ,
t1 = 2 s , and
t2 = 4 s .
Current in a Conductor 01
27:01, calculus, numeric, > 1 min, normal.
001 (part 1 of 2) 1 points
The current
I = a t2 − b t + c
Since
in a section of a conductor depends on time.
What quantity of charge moves across the
section of the conductor from t = 0 to t = t1 ?
1. q =
a t31
¢
¢
¡
2
2 C/s3
3
C/s
q2 =
(4 s)3 −
(4 s)2
3
2
+ (7 C/s)(4 s)
= 46.6667 C , and
¡
2. q = a t21 − b t1 + c
3. q = a t31 − b t2 + c t1
a 3 b 2
t − t + c t1 correct
3 1 2 1
a
b
5. q = t31 − t21 + c
3
2
Explanation:
The unit of current is Coulomb per second
dq
I=
or dq = I dt.
dt
To find the total charge in coulombs that
passes through the conductor, one must integrate the current over the time interval.
4. q =
q=
Z
Z
t1
dq =
0
Z
¡
¢
¢
3 C/s2
2 C/s3
3
(2 s) −
(2 s)2
q1 =
3
2
+ (7 C/s)(2 s)
= 13.3333 C .
¡
so that
q21 = q2 − q1 = 33.3333 C .
keywords:
t1
I dt
0
t1
(a t2 − b t + c) dt
0
·
¸ ¯ t1
¯
a 3 b 2
=
t1 − t1 + c t ¯¯
3
2
0
a 3 b 2
= t1 − t1 + c t 1 .
3
2
=
¸ ¯ t2
¯
a 3 b 2
q=
t − t + c t ¯¯ ,
3
2
t1
·
then
b
− t21 + c t1
2
1
002 (part 2 of 2) 1 points
If I is in A, and a = 2 C/s3 , b = 3 C/s2 ,
and c = 7 C/s, what quantity of charge moves
across the section of the conductor from t1 =
Drift Speed in Copper Wire
27:02, trigonometry, numeric, > 1 min, normal.
003 (part 1 of 1) 2 points
Calculate the average drift speed of electrons
traveling through a copper wire with a crosssectional area of 5.261 mm2 when carrying a
current of 20 A (values similar to those for the
electric wire to your study lamp). Assume
one electron per atom of copper contributes
to the current. The atomic mass of copper
is 63.5 g/mol and its density is 8.93 g/cm3 .
Avogadro’s number NA is 6.02 × 1023 .
Correct answer: 0.000280652 m/s.
Version One – Homework 5 – Savrasov – 39821 – Apr 16, 2007
Explanation:
Let : N = 1 ,
M = 63.5 g/mol ,
ρ = 8.93 g/cm3 ,
A = 5.261 mm2
= 5.261 × 10−6 m2 ,
I = 20 A , and
qe = 1.6 × 10−19 C/electron .
We first calculate n, the number of currentcarrying electrons per unit volume in copper.
Assuming one free conduction electron per
NA ρ
atom, n =
, where NA is Avogodro’s
M
number and ρ and M are the density and the
atomic weight of copper, respectively
¶
µ
electron NA ρ
.
n≡ 1
atom
M
¶
electron 6.02 × 1023 atoms
n= 1
atom
63.5 g/mol
µ 6
¶
¡
¢ 10 cm3
3
× 8.93 g/cm
1 m3
= 8.46592 × 1028 electrons/m3 .
µ
The current in a resistor decreases by 3 A
when the voltage applied across the resistor
decreases from 12 V to 6 V.
Find the resistance of the resistor.
Correct answer: 2 Ω.
Explanation:
Let : ∆I = 3 A ,
Vi = 12 V ,
Vf = 6 V .
=
Call the initial voltage and current Vi and I,
respectively, and the final voltage and current
Vf and (I − ∆I), respectively.
Vf
V
Vi
=
=
I
I
I − ∆I
Vi (I − ∆I) = Vf I
(Vi − Vf ) I = Vi ∆I
Vi − V f
Vi
=
,
I
∆I
R=
so
Vi − V f
Vi
=
I
∆I
12 V − 6 V
=
3A
= 2Ω .
R=
I
n qe A
20 A
8.46592 × 1028 electrons/m3
1
×
1.6 × 10−19 C/electron
1
×
5.261 × 10−6 m2
= 0.000280652 m/s .
keywords:
Resistance
27:04, trigonometry, numeric, > 1 min, normal.
004 (part 1 of 1) 2 points
and
V =IR
Drift Speed in Copper Wire
vd =
2
keywords:
Serway CP 17 11
27:04, trigonometry, numeric, > 1 min, normal.
005 (part 1 of 2) 1 points
A person notices a mild shock if the current
along a path through the thumb and index
finger exceeds 80 µA.
Find the maximum allowable potential
difference without shock across the thumb
and index finger for a dry-skin resistance of
400000 Ω.
Correct answer: 32 V.
Explanation:
Version One – Homework 5 – Savrasov – 39821 – Apr 16, 2007
3
The area is
Let : I = 80 µA = 8 × 10
R = 400000 Ω .
−5
A
A = π r2
and
and the resistance is
L
L
= ρ20
A
π r2
−8
= (1.7 × 10 Ω · m)
15 m
×
π (0.000512 m)2
The potential difference for dry skin is
R = ρ20
∆V = I R
= (8 × 10−5 A) (400000 Ω)
= 32 V .
= 0.309635 Ω .
006 (part 2 of 2) 1 points
Find the maximum allowable potential difference without shock across the thumb and index finger for a wet-skin resistance of 2000 Ω.
Correct answer: 0.16 V.
Explanation:
Let : R = 2000 Ω .
008 (part 2 of 2) 1 points
Calculate the difference in the resistance of
the wire between 45◦ C and 60◦ C.
Correct answer: 0.0176492 Ω.
Explanation:
The potential difference for wet skin is
Let :
∆V = I R
= (8 × 10−5 A) (2000 Ω)
= 0.16 V .
T1 = 45◦ C ,
T2 = 60◦ C ,
r = 0.512 mm = 0.000512 m ,
ρ20 = 1.7 × 10−8 Ω · m ,
α = 0.0038 (◦ C)−1 , and
L = 15 m .
keywords:
Wire Resistance vs Temperature
27:06, trigonometry, numeric, > 1 min, normal.
007 (part 1 of 2) 1 points
A 15 m long piece of wire of density 8.34 g/m3
has a diameter of 1.024 mm. The resistivity of the wire is 1.7 × 10−8 Ω · m at 20◦ C.
The temperature coefficient for the wire is
0.0038 (◦ C)−1 .
Calculate the resistance of the wire at
20◦ C.
Correct answer: 0.309635 Ω.
Explanation:
Let : L = 15 m ,
r = 0.512 mm = 0.000512 m ,
ρ20 = 1.7 × 10−8 Ω · m .
and
A = π r2 ,
and
ρ1 = ρ20 {1.0 + α [T1 − T20 ]}
= (1.7 × 10−8 Ω · m)
× {1.0 + [0.0038 (◦ C)−1 ]
× [(45◦ C) − (20◦ C)]}
= 1.8615 × 10−8 Ω · m .
L
L
= ρ1
A
π r2
= (1.8615 × 10−8 Ω · m)
15 m
×
π (0.000512 m)2
= 0.339051 Ω .
R1 = ρ 1
Version One – Homework 5 – Savrasov – 39821 – Apr 16, 2007
ρ2 = ρ20 {1.0 + α [T2 − T20 ]}
= (1.7 × 10−8 Ω · m)
× {1.0 + [0.0038 (◦ C)−1 ]
× [(60◦ C) − (20◦ C)]}
= 1.9584 × 10−8 Ω · m .
L
L
= ρ2
A
π r2
= (1.9584 × 10−8 Ω · m)
15 m
×
π (0.000512 m)2
= 0.3567 Ω .
R2 = ρ 2
So the difference in the resistance is
∆R = |R2 − R1 |
= |0.3567 Ω − 0.339051 Ω|
= 0.0176492 Ω .
keywords:
4