GenericStudent – Homework 5 – savrasov – 39823 – Jan 02, 2008
This print-out should have 5 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. The due time is Central
time.
001 (part 1 of 1) 2 points
Calculate the average drift speed of electrons
traveling through a copper wire with a crosssectional area of 40 mm2 when carrying a
current of 5 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: 9.22817 × 10−6 m/s.
Explanation:
Let : N = 1 ,
M = 63.5 g/mol ,
ρ = 8.93 g/cm3 ,
A = 40 mm2
= 4 × 10−5 m2 ,
I = 5 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
, where NA is Avogodro’s
atom, n =
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
Drift Speed in Copper Wire
I
vd =
n qe A
=
1
5A
8.46592 × 1028 electrons/m3
1
×
−19
1.6 × 10
C/electron
1
×
4 × 10−5 m2
= 9.22817 × 10−6 m/s .
keywords:
002 (part 1 of 1) 2 points
A wire is made of a material with a resistivity of 3.13928 × 10−8 Ω · m. It has length
2.36431 m and diameter 0.73598 mm.
What is the resistance of the wire?
Correct answer: 0.174467 Ω.
Explanation:
Let :
ρ = 3.13928 × 10−8 Ω · m ,
` = 2.36431 m , and
r = 0.36799 mm = 0.00036799 m .
By definition, the resistance of the wire is
`
ρ`
R=ρ =
π r2
¡A
¢
3.13928 × 10−8 Ω · m (2.36431 m)
=
π (0.00036799 m)2
= 0.174467 Ω .
keywords:
003 (part 1 of 1) 2 points
A 0.62 V potential difference is maintained
across a 1.8 m length of tungsten wire that
has a cross-sectional area of 0.4 mm2 and the
resistivity of the tungsten is 5.6 × 10−8 Ω · m.
What is the current in the wire?
Correct answer: 2.46032 A.
Explanation:
Let :
V = 0.62 V ,
` = 1.8 m ,
A = 0.4 mm2 = 4 × 10−7 m2 ,
ρ = 5.6 × 10
−8
Ω · m.
and
GenericStudent – Homework 5 – savrasov – 39823 – Jan 02, 2008
2
The resistance is
V
ρ`
R=
=
,
I
A
so the current is
VA
ρ`
(0.62 V) (4 × 10−7 m2 )
=
(5.6 × 10−8 Ω · m) (1.8 m)
I=
= 2.46032 A .
005 (part 1 of 1) 2 points
At 32 ◦ C the carbon resistor in an electric
circuit, connected to a 4.3 V battery, has a
resistance of 162 Ω.
What is the current in the circuit when the
temperature of the carbon rises to 71◦ C? Use
a temperature coefficient of −0.0005 (◦ C)−1 .
Correct answer: 27.0711 mA.
Explanation:
Let :
keywords:
004 (part 1 of 1) 2 points
One wire of a 13 m extension cord made of
16-gauge copper wire carrying a current of
3 A.
What is the potential difference across the
cord?
Correct answer: 0.0947143 V.
Explanation:
T0
V
R0
T1
= 32◦ C ,
= 4.3 V ,
= 162 Ω ,
= 71◦ C , and
α = −0.0005 (◦ C)−1 .
At 71◦ C the resistance has increased to
R = R0 (1 + α ∆T )
= (162 Ω)
©
£
¤
ª
× 1 + −0.0005 (◦ C)−1 (39◦ C)
= 158.841 Ω ,
so the current is
Let : ρ = 1.7 × 10
` = 13 m ,
−8
Ω · m,
A = 7 mm2
= 7 × 10−6 m2 ,
I = 3 A.
V
4.3 V
103 mA
=
·
R
158.841 Ω
1A
= 27.0711 mA .
I=
and
keywords:
The resistance is
ρ`
A
(1.7 × 10−8 Ω · m) (13 m)
=
7 × 10−6 m2
= 0.0315714 Ω .
R=
By Ohm’s Law,
V = RI
= (0.0315714 Ω) (3 A)
= 0.0947143 V .
keywords: