Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 21, Problem 76P

(a)

To determine

The resistance of the wires and the corresponding values of the resistivity.

(a)

Expert Solution
Check Mark

Answer to Problem 76P

The resistance of wire 1 is 7.25Ω , the resistance of wire 2 is 14.06Ω , the resistance of wire 3 is 21.14Ω , the resistivity of wire 1 is 9.80×107Ωm , the resistivity of wire 2 is 9.98×107Ωm , the resistivity of wire 3 is 1.00×106Ωm .

Explanation of Solution

Given information: Area of cross section of gauge wire is 7.30×108m2 , length of gauge wire 1 is 0.540m , length of gauge wire 2 is 1.028m , length of gauge wire 3 is 1.543m , voltage across the wire 1 is 5.22V , voltage across the wire 2 is 5.82V , voltage across the wire 3 is 5.94V , current across the wire 1 is 0.72A , current across the wire 2 is 0.414A , current across the wire 3 is 0.281A .

Formula to calculate the resistance of wire 1.

V1=I1R1R1=V1I1 (1)

Here,

R1 is the resistance of wire 1.

I1 is the current across the wire 1.

V1 is the voltage across the wire 1.

Substitute 5.22V for V1 , 0.72A for I1 in equation (1) to find R1 ,

R1=5.22V0.72A=7.25Ω

Thus, the resistance of wire 1 is 7.25Ω .

Formula to calculate the resistivity of wire 1.

R1=ρ1l1Aρ1=R1×Al1 (2)

Here,

l1 is the the length of the wire 1.

ρ1 is the resistivity of wire 1.

A is the cross section of the gauge wire.

Substitute 7.25Ω for R1 , 0.540m for l1 , 7.30×108m2 for A in equation (2) to find ρ1 .

ρ1=7.25Ω×7.30×108m20.540m=9.80×107Ωm

Thus, the resistivity of wire 1 is 9.80×107Ωm .

Formula to calculate the resistance of wire 2.

V2=I2R2R2=V2I2 (3)

Here,

R2 is the resistance of wire 2.

I2 is the current across the wire 2.

V2 is the voltage across the wire 2.

Substitute 5.82V for V2 , 0.414A for I2 in equation (3) to find R2 ,

R1=5.82V0.414A=14.057Ω14.06Ω

Thus, the resistance of wire 2 is 14.06Ω .

Formula to calculate the resistivity of wire 2.

R2=ρ2l2Aρ2=R2×Al2 (4)

Here,

l2 is the the length of the wire 2.

ρ2 is the resistivity of wire 2.

A is the cross section of the gauge wire.

Substitute 14.06Ω for R2 , 1.028m for l2 , 7.30×108m2 for A in equation (4) to find ρ2 ,

ρ2=14.06Ω×7.30×108m21.028m=9.98×107Ωm

Thus, the resistivity of wire 2 is 9.98×107Ωm .

Formula to calculate the resistance of wire 3.

V3=I3R3R3=V3I3 (5)

Here,

R3 is the resistance of wire 3.

I3 is the current across the wire 3.

V3 is the voltage across the wire 3.

Substitute 5.94V for V3 , 0.281A for I3 in equation (5) to find R3 ,

R3=5.94V0.281A=21.138Ω21.14Ω

Thus, the resistance of wire 3 is 21.14Ω .

Formula to calculate the resistivity of wire 3.

R3=ρ3l3Aρ3=R3×Al3 (6)

Here,

l3 is the the length of the wire 3.

ρ3 is the resistivity of wire 3.

A is the cross section of the gauge wire.

Substitute 21.14Ω for R3 , 1.543m for l3 , 7.30×108m2 for A in equation (6) to find ρ3 ,

ρ3=21.14Ω×7.30×108m21.543m=1.00×106Ωm

Thus, the resistivity of wire 3 is 1.00×106Ωm .

Conclusion:

Therefore, the resistance of wire 1 is 7.25Ω , the resistance of wire 2 is 14.06Ω , the resistance of wire 3 is 21.14Ω , the resistivity of wire 1 is 9.80×107Ωm , the resistivity of wire 2 is 9.98×107Ωm , the resistivity of wire 3 is 1.00×106Ωm .

(b)

To determine

The average value of resistivity.

(b)

Expert Solution
Check Mark

Answer to Problem 76P

The average value of resistivity is 1.0×106Ωm .

Explanation of Solution

Given information: Area of cross section of gauge wire is 7.30×108m2 , length of gauge wire 1 is 0.540m , length of gauge wire 2 is 1.028m , length of gauge wire 3 is 1.543m , voltage across the wire 1 is 5.22V , voltage across the wire 2 is 5.82V , voltage across the wire 3 is 5.94V , current across the wire 1 is 0.72A , current across the wire 2 is 0.414A , current across the wire 3 is 0.281A .

Formula to calculate the average value of resistivity.

ρavg=ρ1+ρ2+ρ33 (7)

Here,

ρavg is the average value of resistivity.

Substitute 9.80×107Ωm for ρ1 , 9.98×107Ωm for ρ2 , 1.00×106Ωm for ρ3 in equation (7) to find ρavg ,

ρavg=(9.80×107Ωm)+(9.98×107Ωm)+(1.00×106Ωm)3=0.9926×106Ωm1.0×106Ωm

Thus, the average value of resistivity is 1.0×106Ωm .

Conclusion:

Therefore, the average value of resistivity is 1.0×106Ωm .

(c)

To determine

The way in which this average value of resistivity compares with the value given in the table 26.2.

(c)

Expert Solution
Check Mark

Answer to Problem 76P

This average value of resistivity compares with the value given in the table 26.2. It is observed that the average value of the resistivity is equal to the resistivity of the nichrome. So, a experiment is done by the student to measure the electrical resistance of Nichrome in the form of wires of different lengths and cross sectional areas is correct.

Explanation of Solution

Given information: Area of cross section of gauge wire is 7.30×108m2 , length of gauge wire 1 is 0.540m , length of gauge wire 2 is 1.028m , length of gauge wire 3 is 1.543m , voltage across the wire 1 is 5.22V , voltage across the wire 2 is 5.82V , voltage across the wire 3 is 5.94V , current across the wire 1 is 0.72A , current across the wire 2 is 0.414A , current across the wire 3 is 0.281A .

The average value of resistivity is 1.0×106Ωm .

The value of resistivity of Nichrome is ranging from 1.00×106Ωmto1.50×106Ωm .

It is observed that the average value of the resistivity is equal to the resistivity of the nichrome. So, a experiment is done by the student to measure the electrical resistance of Nichrome in the form of wires of different lengths and cross sectional areas is correct.

Thus, this average value of resistivity compares with the value given in the table 26.2. It is observed that the average value of the resistivity is equal to the resistivity of the nichrome. So, a experiment is done by the student to measure the electrical resistance of Nichrome in the form of wires of different lengths and cross sectional areas is correct.

Conclusion:

Therefore, this average value of resistivity compares with the value given in the table 26.2. It is observed that the average value of the resistivity is equal to the resistivity of the nichrome. So, a experiment is done by the student to measure the electrical resistance of Nichrome in the form of wires of different lengths and cross sectional areas is correct.

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Principles of Physics: A Calculus-Based Text

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