This preview shows page 1. Sign up to view the full content.
Unformatted text preview: _________________ 1064 ELECTRIC CIRCUITS 17. REASONING Assuming that the resistance is R at a temperature T and R0 at a temperature
T0, we can write the percentage change p in resistance as p= R − R0
R0 × 100 Equation 20.5, on the other hand, gives the resistance as a function of temperature as follows:
R = R0 1 + α (T − T0 ) where α is the temperature coefficient of resistivity. Substituting this expression into the
expression for the percentage change in resistance gives
p= R − R0
R0 × 100 = R0 + R0α (T − T0 ) − R0
R0 ×100 = α (T − T0 ) 100 (1) The change in temperature is unknown, but it is the same for both wires. Therefore, we will
apply Equation (1) to each wire and divide the two expressions to eliminate the unknown
change in temperature. From the result we will be able to calculate the percentage change in
the resistance of the tungsten wire.
SOLUTION Applying Equation (1) to both wires gives pTungsten = α Tungsten (T − T0 )100 and pGold = α Gold (T − T0 )100 Dividing these two expressions, eliminating (T − T0) algebraically, and solving for pTungsten
give
pTungsten α Tungsten (T − T0 )100 α Tungsten
=
=
α Gold
pGold
α Gold (T − T0 )100 pTungsten α Tungsten
= pGold α Gold −1 0.0045 ( C° )
= 9.3% = ( 7.0% )
−1 0.0034 ( C° ) 18. REASONING AND SOLUTION Suppose that when the initial temperature of the wire is
T0 the resistance is R0, and when the temperature rises to T the resistance is R. The relation
between temperature and resistance is given by Equation 20.5 as R = R0[1 + α (T – T0)], where α is the temperature coefficient of resistivity. The initial and final resistances are
related to the voltage and current as R0 = V/I0 and R = V/I, where the voltage V across the Chapter 20 Problems 1065 wire is the same in both cases. Substituting these values for R0 and R into Equation 20.5 and
solving for T, we arrive at T = T0 + I0 − 1 I = 360 ° C
4.5 × 10 −4 (C° ) −1
______________________________________________________________________________
19. α = 20 ° C + 1.50 A − 1 1.30 A 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 . The volume V2 of the new
wire will be the same as the original volume V1 of the wire, where volume is the product of
length and crosssectional area. Thus, V1 = V2 or A1L1 = A2L2. Since the new wire is three
times longer than the first wire, we can write
A1L1 = A2 L2 = A2 (3L1) or A2 = A1 / 3 We can form the ratio of the resistances, use this expression for the area A2 , and find the
new resistance.
SOLUTION The resistance of the new wire is determined as follows:
R2
R1 = ρ L2 / A2 L2 A1 ( 3L1 ) A1
=
=
=9
ρ L1 / A1 L1 A2 L1 ( A1 / 3) Solving for R2 , we find that
R2 = 9 R1 = 9(21.0 Ω ) = 189 Ω ______________________________________________________________________________
20. REASONING The length L of the wire is related to its resistance R and crosssectional area
A by L = A R / ρ (see Equation 20.3), where ρ is the resistivity of tungsten. The resistivity
is known (see Table 20.1), and the crosssectional area can be determined since the radius of
the wire is given. The resistance can be obtained from Ohm’s law as the voltage divided by
the current.
SOLUTION The length L of the wire is
L= AR ρ (1) 1066 ELECTRIC CIRCUITS Since the crosssection of the wire is circular, its area is A = π r 2 , where r is the radius of
the wire. According to Ohm’s law (Equation 20.2), the resistance R is related to the voltage
V and current I by R = V / I . Substituting the expressions for A and R into Equation (1)
gives (π r 2 ) V 2 120 V (
−3 π 0.0030 ×10 m ) AR
I= 1.24 A = 0.049 m
L=
=
ρ
ρ
5.6 ×10−8 Ω ⋅ m
______________________________________________________________________________ 21. REASONING AND SOLUTION The mass m of the aluminum wire is equal to the density
d of aluminum times the volume of the wire. The wire is cylindrical, so its volume is equal
to the crosssectional area A times the length L; m = dAL.
The crosssectional area of the wire is related to its resistance R and length L by
Equation 20.3; R = ρL/A, where ρ is the resistivity of aluminum (see Table 20.1).
Therefore, the mass of the aluminum wire can be written as ρL L
m = dAL = d R
The resistance R is given by Ohm’s law as R = V/I, so the mass of the wire becomes d ρ L2 I ρL m=d
L= V
R
(2700 kg/m3 )(2.82 × 10−8 Ω ⋅ m)(175 m) 2 (125 A)
= 9.7 × 102 kg
0.300 V
______________________________________________________________________________
m= 22. REASONING AND SOLUTION According to Equation 20.6c, the power delivered to the
iron is
2
V 2 (120 V)
P=
=
= 6.0 × 10 2 W
R
24 Ω
______________________________________________________________________________ 23. REASONING AND SOLUTION
a. According to Equation 20.6c, the resistance is
2
V 2 (12 V )
R=
=
= 4.4 Ω
P
33 W Chapter 20 Problems 1067 b. According to Equation 20.6a, the current is P 33 W
=
= 2.8 A...
View
Full
Document
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

Click to edit the document details