EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100546310
Author: Jewett
Publisher: CENGAGE L
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Chapter 27, Problem 27.75AP

Review. When a straight wire is warmed, its resistance is given by R = R0,[1 + a(TT0)] according to Equation 27.20, where a is the temperature coefficient of resistivity. This expression needs to be modified if we include the change in dimensions of the wire due to thermal expansion. For a copper wire of radius 0.100 0 mm and length 2.000 m, find its resistance at 100.0°C, including the effects of both thermal expansion and temperature variation of resistivity. Assume the coefficients are known to four significant figures.

Expert Solution & Answer
Check Mark
To determine

The resistance of wire at 100°C .

Answer to Problem 27.75AP

The resistance of wire at 100°C is 1.418Ω .

Explanation of Solution

Given info: The radius of copper wire is 0.100mm and length is 2.000m . The temperature coefficient of copper is 3.9×103(°C)1 . The resistance of a straight wire is R=R0[1+α(TT0)] .

Write the expression of the resistance of the wire.

R=R0[1+α(TT0)]

Here,

R is the resistance of the wire.

R0 is the resistance of initial temperature.

α is the temperature coefficient of resistivity.

T is the increased temperature

T0 is the initial temperature.

The resistance of wire is increased with the increase of temperature because increase in temperature causes to collision between free electrons in metal with crystal lattice ions.

The length of wire is changed with the increase of temperature. So the final length of the wire is,

l=l0(1+α(TT0))

Here,

l is the length of the wire at temperature T .

l0 is the length of wire at temperature T0 .

α is the coefficient of linear expansion.

The area of the cross section of the wire at temperature T is,

A=A0(1+2α(TT0))

Here,

A is the area of the cross section of wire at temperature T .

A0 is the area of the cross section of wire at temperature T0 .

Resistivity of wire with change in temperature is,

ρ=ρ0(1+α(TT0))

Here,

ρ is the resistivity of wire at temperature T .

ρ0 is the resistivity of wire at temperature T0 .

The resistivity of copper is 1.7×108Ωm .

Write the expression for the resistance of wire at 100°C .

R0=ρ0l0A0 (1)

The area of wire is,

A0=πr2

Here,

r is the radius of wire.

Substitute 0.100mm for r in above equation.

A0=π(0.100mm(1×103m1mm))2=π(0.100×103m)2=0.0314×106m2

Thus, the area of the wire is 0.0314×106m2 .

Substitute 1.7×108Ωm for ρ0 , 2.000m for l0 and 0.0314×106m2 for A0 in equation (1).

R0=1.7×108Ωm×(2.000m0.0314×106m2)=(1.7×108Ωm)(63.69×106m1)=108.273×102Ω

The relation between resistance of wire, the length of wire and the cross sectional area at final temperature is,

R=ρlA

Substitute ρ0(1+α(TT0)) for ρ , l0(1+α(TT0)) for l and A0(1+α(TT0)) for A in above equation.

R=ρ0(1+α(TT0))l0(1+α(TT0))A0(1+2α(TT0))=ρ0l0A0×[1+α(TT0)](1+α(TT0))(1+2α(TT0))

Substitute R0 for ρ0l0A0 in above equation.

R=R0×[1+α(TT0)](1+α(TT0))(1+2α(TT0))

The coefficient of linear expansion is 17×106(°C)1 .

Substitute 108.273×102Ω for R0 , 17×106(°C)1 for α , 3.9×103(°C)1 for α , 20°C for T0 and 100°C for T in above equation.

R=[108.273×102Ω×[1+(3.9×103(°C)1)(10020)°C](1+(17×106(°C)1)(10020)°C)](1+2(17×106(°C)1)(10020)°C)=108.273×102Ω×[1.312](1.00136)(1.00272)=1.418Ω

Conclusion:

Therefore, the resistance of wire at 100°C is 1.418Ω .

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Chapter 27 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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