Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 27, Problem 62AP

(a)

To determine

The resistance and resistivity of the wire in each case.

(a)

Expert Solution
Check Mark

Answer to Problem 62AP

The resistance and resistivity of the wire in each case is given in the table below.

L(m)ΔV(V)I(A)R(Ω)ρ(Ωm)
0.5405.220.727.259.8×107
1.0285.820.41414.11.0×106
1.5435.940.28121.11.0×106

Explanation of Solution

Write the expression to obtain the resistance of the wire.

    R=ΔVI                                                                                                                    (I)

Here, R is the resistance, ΔV is the voltage and I is the current.

Write the expression to obtain the resistivity of the wire.

    ρ=RAL                                                                                                                  (II)

Here, ρ is the resistivity, R is the resistance, A is the area of cross section of the wire and L is the length of the wire.

Conclusion:

Case (i):

Substitute 5.22V for ΔV and 0.72A for I in equation (I) to calculate R.

    R=5.22V0.72A=7.25Ω

Substitute 7.25Ω for R, 0.540m for L and 7.30×108m2 for A in equation (II) to calculate ρ.

    ρ=(7.25Ω)7.30×108m20.540m=9.8×107Ωm

Case (ii):

Substitute 5.82V for ΔV and 0.414A for I in equation (I) to calculate R.

    R=5.82V0.414A=14.1Ω

Substitute 14.1Ω for R, 1.028m for L and 7.30×108m2 for A in equation (II) to calculate ρ.

    ρ=(14.1Ω)7.30×108m21.028m=1.0×106Ωm

Case (iii):

Substitute 5.94V for ΔV and 0.281A for I in equation (I) to calculate R.

    R=5.94V0.281A=21.1Ω

Substitute 21.1Ω for R, 1.543m for L and 7.30×108m2 for A in equation (II) to calculate ρ.

    ρ=(21.1Ω)7.30×108m21.543m=1.0×106Ωm

Therefore, the resistance and resistivity of the wire in each case is given in the table below.

L(m)ΔV(V)I(A)R(Ω)ρ(Ωm)
0.5405.220.727.259.8×107
1.0285.820.41414.11.0×106
1.5435.940.28121.11.0×106

(b)

To determine

The average resistivity.

(b)

Expert Solution
Check Mark

Answer to Problem 62AP

The average resistivity is 0.993×106Ωm.

Explanation of Solution

Write the expression to obtain the average resistivity.

    ρavgρ1+ρ2+ρ33

Here, ρavg is the average resistivity, ρ1 is the resistivity of material 1, ρ2 is the resistivity of material 2 and ρ3 is the resistivity of material 3.

Conclusion:

Substitute 9.8×107Ωm for ρ1. 1.0×106Ωm for ρ2 and 1.0×106Ωm for ρ3 in the above equation to calculate ρavg.

    ρavg=(9.8×107Ωm)+(1.0×106Ωm)+(1.0×106Ωm)3=2.98×106Ωm3=0.993×106Ωm

Therefore, the average resistivity is 0.993×106Ωm.

(c)

To determine

The comparison between the average resistivity and the resistivity values for different length as shown in table in part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 62AP

There is not much variation in the resistivity value for different lengths for the same material.

Explanation of Solution

The values of the resistivity with different length and cross section is as below.

L(m)ΔV(V)I(A)R(Ω)ρ(Ω-m)
0.5405.220.727.259.8×107
1.0285.820.41414.11.0×106
1.5435.940.28121.11.0×106

The average value of resistivity is 0.993×106Ω-m.

The values of the resistivity in the above table and the average value are almost similar.

Conclusion:

Therefore, there is not much variation in the resistivity value for different length for the same material.

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Chapter 27 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

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