Unformatted text preview: terial. In order to rank
the resistances, we need to evaluate L and A for each configuration in terms of L0, the unit
of length.
Resistance
a
b
c 2
= ρ L L0 × 2 L0 0
1
L0
R=ρ
= ρ 8L 2 L0 × 4 L0 0
1
2 L0
R=ρ
= ρ 2L L0 × 4 L0 0
R=ρ 4 L0 Rank
1
3
2 Therefore, we expect that a has the largest resistance, followed by c, and then by b.
b. Equation 20.2 states that the current I is equal to the voltage V divided by the resistance,
I = V / R . Since the current is inversely proportional to the resistance, the largest current
arises when the resistance is smallest, and vice versa. Thus, we expect that b has the largest
current, followed by c, and then by a.
SOLUTION
a. The resistances can be found by using the results from the REASONING: Chapter 20 Problems a
b
c 1061 2 2
−2
R = ρ = 1.50 × 10 Ω ⋅ m = 0.600 Ω
−2
L 5.00 × 10 m 0
1 1
−2
R = ρ = 1.50 × 10 Ω ⋅ m = 0.0375 Ω
−2 8L 8 × 5.00 × 10 m 0
1 1
−2
R = ρ = 1.50 × 10 Ω ⋅ m = 0.150 Ω
−2 2L 0 2 × 5.00 × 10 m ( ) ( ) ( ) b. The current in each case is given by Equation 20.2, where the value of the resistance is
obtained from part (a):
3.00 V
V
I= =
= 5.00 A
a
R 0.600 Ω
3.00 V
V
I= =
= 80.0 A
b
R 0.0375 Ω
3.00 V
V
I= =
= 20.0 A
c
R 0.150 Ω
______________________________________________________________________________ 11. REASONING The resistance R of a wire that has a length L and a crosssectional area A is
L
given by Equation 20.3 as R = ρ . Both wires have the same length and crosssectional
A
area. Only the resistivity ρ of the wire differs, and Table 20.1 gives the following values:
ρAluminum = 2.82 × 10−8 ·m and ρCopper = 1.72 × 10−8 ·m. Applying Equation 20.3 to both
wires and dividing the two equations will allow us to eliminate the unknown length and
crosssectional area algebraically and solve for the resistance of the copper wire.
SOLUTION Applying Equation 20.3 to both wires gives RCopper = ρCopper L
A and RAluminum = ρ Aluminum L
A Dividing these two equations, eliminating L and A algebraically, and solving the result for
RCopper give
RCopper
ρ Copper L / A
ρ Copper
=
=
RAluminum ρ Aluminum L / A ρ Aluminum ρCopper 1.72 ×10−8 Ω ⋅ m RCopper = RAluminum = ( 0.20 Ω ) = 0.12 Ω
−8 2.82 ×10 Ω ⋅ m ρ Aluminum 1062 ELECTRIC CIRCUITS 12. REASONING AND SOLUTION The resistance of the cable is R= V ρL
=
I
A Since A = π r2, the radius of the cable is (1.72 × 10–8 Ω ⋅ m ) ( 0.24 m )(1200 A ) =
π (1.6 × 10 –2 V ) ρ LI
=
r=
πV 9.9 × 10 –3 m ______________________________________________________________________________
13. REASONING AND SOLUTION From Equation 20.5 we have that R = R0[1 + α (T – T0)].
Solving for T gives
99.6 Ω
−1
125 Ω
= 20.0 °C +
= −34.6 °C
T = T0 +
α
3.72 × 10−3 (C°) −1
______________________________________________________________________________
R
−1
R0 14. REASONING AND SOLUTION Using Equation 20.3 and the resistivity of aluminum
from Table 20.1, we find R= ρL ( 2.82 × 10–8 Ω ⋅ m ) (10.0 × 103 m ) =
= 0.58 Ω
A
4.9 × 10 –4 m 2
______________________________________________________________________________
15. SSM WWW REASONING The resistance of a metal wire of length L, crosssectional
area A and resistivity ρ is given by Equation 20.3: R = ρ L / A . Solving for A, we have
A = ρ L / R . We can use this expression to find the ratio of the crosssectional area of the
aluminum wire to that of the copper wire.
SOLUTION Forming the ratio of the areas and using resistivity values from Table 20.1, we
have
Aaluminum ρ aluminum L / R ρ aluminum 2.82 × 10 –8 Ω ⋅ m
=
=
=
= 1.64
ρcopper L / R
ρ copper
Acopper
1.72 × 10 –8 Ω ⋅ m ______________________________________________________________________________ Chapter 20 Problems 1063 16. REASONING The resistance R of the spooled wire decreases as its length L decreases,
L
according to R = ρ (Equation 20.3). The resistivity ρ and crosssectional area A of the
A
wire do not change. Because the same battery is used, the potential difference V across the
V
wire is the same in both cases. Therefore, R = (Equation 20.2) explains why the current I
I
increases as the resistance R of the wire decreases.
RA
L
(Equation 20.3) for L, we obtain L =
. Thus, the initial
A
ρ
length L0 and final length Lf of the wire are given by SOLUTION Solving R = ρ L0 = R0 A ρ and Lf = Rf A ρ (1) where R0 is the initial resistance, and Rf the final resistance, of the wire. Taking the ratio of Equations (1) eliminates the unknown quantities A and ρ, allowing us to solve for Lf in
terms of L0 and the initial and final resistances: Rf A ρ
Lf
R
=
=f
L0 R0 A R0 or ρ From R = R Lf = f L0
R 0 (2) V
(Equation 20.2), the initial resistance R0 and final resistance Rf of the spooled
I wire are
R0 = V
I0 and Rf = V
If (3) Substituting Equations (3) into Equation (2) yields
V I
Lf = f
V I
0 L = I 0 L = 2.4 A 75 m = 58 m
) ( 0 I f 0 3.1 A _____________________________________________________________...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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