Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 27, Problem 32P

(a)

To determine

Weather the design goal is achieved with this method or not.

(a)

Expert Solution
Check Mark

Answer to Problem 32P

The design goal is achieved with this method.

Explanation of Solution

Write the expression to obtain the resistance in the wire.

    R=ρlA

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

Substitute πr2 for A in the above equation.

    R=ρlπr2                                                                                                                     (I)

Here, r is the radius of the wire.

Write the expression to obtain the carbon resistance in the wire.

    R1=ρ1l1πr2                                                                                                                  (II)

Here, R1 is the carbon resistance in the wire, ρ1 is the carbon resistivity, l1 is the length of the carbon wire and r is the radius of the wire.

Write the expression to obtain the nichrome resistance in the wire.

    R2=ρ2l2πr2                                                                                                                 (III)

Here, R2 is the nichrome resistance in the wire, ρ2 is the nichrome resistivity, l2 is the length of the nichrome wire and r is the radius of the wire.

Write the expression to obtain the variation in resistance with change in temperature.

    RT=R(1+αΔT)                                                                                                    (IV)

Here, RT is the resistance with temperature change, R is the initial resistance, α is the coefficient of thermal expansion and ΔT is the change in temperature.

Write the expression to obtain the final resistance.

    R1T+R2T=R1(1+α1ΔT)+R2(1+α2ΔT)                                                               (V)

Here, R1T is carbon resistance variation with temperature, R2T is nichrome resistance variation with temperature, R1 is the initial carbon resistance, R2 is the initial nichrome resistance, α1 is the coefficient of thermal resistance for carbon resistance, α2 is the coefficient of thermal resistance for nichrome resistance and ΔT is the change in temperature.

Resistance is independent of temperature variation, thus,

    R1T+R2T=R1+R2                                                                                                 (VI)

Compare equation (III) and (IV).

    R1α1ΔT+R2α2ΔT=0R1R2=α2α1

Substitute 0.5×103°C1 for α1 and 0.4×103°C1 for α2 in the above equation.

    R1R2=0.4×103°C1(0.5×103°C1)R1R2=0.8R1=0.8R2                                                                                      (VII)

The sum of the resistances is given as 10.0Ω.

    R1+R2=10.0Ω                                                                                                    (VII)

Conclusion:

Substitute 0.8R2 for R1 in the above equation to calculate R2.

    0.8R2+R2=10.0Ω1.8R2=10.0ΩR2=10.0Ω1.8R2=5.56Ω

Substitute 5.56Ω for R2 in equation (VII) to calculate R1.

    R1=0.8(5.56Ω)=4.44Ω

As the equation (VII) and (VIII) provides the suitable values of the resistances,

Therefore, the design goal is achieved with this method.

(b)

To determine

The length of the carbon and the nichrome cylinder.

(b)

Expert Solution
Check Mark

Answer to Problem 32P

The length of the carbon cylinder is 0.8967m and length of the nichrome cylinder is 26.2m.

Explanation of Solution

Conclusion:

Substitute 4.44Ω for R1 and 1.50mm for r and 3.50×105Ω-m for ρ1 in equation (II) to calculate l1.

    4.44Ω=(3.50×105Ω-m)l1π(1.50mm)2l1=(4.44Ω)(π(1.50mm)2)(3.50×105Ω-m)l1=(4.44Ω)(π(1.50mm×1m103mm)2)(3.50×105Ω-m)l1=0.8967m

Substitute 5.56Ω for R2 and 1.50mm for r and 1.50×106Ω-m for ρ2  in equation (III) to calculate l2.

    5.56Ω=(1.50×106Ω-m)l2π(1.50mm)2l2=(5.56Ω)(π(1.50mm)2)(1.50×106Ω-m)l2=(5.56Ω)(π(1.50mm×1m103mm)2)(1.50×106Ω-m)l2=26.2m

Therefore, the length of the carbon cylinder is 0.8967m and length of the nichrome cylinder is 26.2m.

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Chapter 27 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 27 - Prob. 6OQCh. 27 - Prob. 7OQCh. 27 - Prob. 8OQCh. 27 - Prob. 9OQCh. 27 - Prob. 10OQCh. 27 - Prob. 11OQCh. 27 - Prob. 12OQCh. 27 - Prob. 13OQCh. 27 - Prob. 1CQCh. 27 - Prob. 2CQCh. 27 - Prob. 3CQCh. 27 - Prob. 4CQCh. 27 - Prob. 5CQCh. 27 - Prob. 6CQCh. 27 - Prob. 7CQCh. 27 - Prob. 8CQCh. 27 - Prob. 1PCh. 27 - A small sphere that carries a charge q is whirled...Ch. 27 - Prob. 3PCh. 27 - Prob. 4PCh. 27 - Prob. 5PCh. 27 - Prob. 6PCh. 27 - Prob. 7PCh. 27 - Prob. 8PCh. 27 - The quantity of charge q (in coulombs) that has...Ch. 27 - Prob. 10PCh. 27 - Prob. 11PCh. 27 - Prob. 12PCh. 27 - Prob. 13PCh. 27 - Prob. 14PCh. 27 - A wire 50.0 m long and 2.00 mm in diameter is...Ch. 27 - A 0.900-V potential difference is maintained...Ch. 27 - Prob. 17PCh. 27 - Prob. 18PCh. 27 - Prob. 19PCh. 27 - Prob. 20PCh. 27 - Prob. 21PCh. 27 - Prob. 22PCh. 27 - Prob. 23PCh. 27 - Prob. 24PCh. 27 - Prob. 25PCh. 27 - Prob. 26PCh. 27 - Prob. 27PCh. 27 - While taking photographs in Death Valley on a day...Ch. 27 - Prob. 29PCh. 27 - Prob. 30PCh. 27 - Prob. 31PCh. 27 - Prob. 32PCh. 27 - Prob. 33PCh. 27 - Prob. 34PCh. 27 - At what temperature will aluminum have a...Ch. 27 - Assume that global lightning on the Earth...Ch. 27 - Prob. 37PCh. 27 - Prob. 38PCh. 27 - Prob. 39PCh. 27 - The potential difference across a resting neuron...Ch. 27 - Prob. 41PCh. 27 - Prob. 42PCh. 27 - Prob. 43PCh. 27 - Prob. 44PCh. 27 - Prob. 45PCh. 27 - Prob. 46PCh. 27 - Prob. 47PCh. 27 - Prob. 48PCh. 27 - Prob. 49PCh. 27 - Prob. 50PCh. 27 - Prob. 51PCh. 27 - Prob. 52PCh. 27 - Prob. 53PCh. 27 - Prob. 54PCh. 27 - Prob. 55PCh. 27 - Prob. 56PCh. 27 - Prob. 57APCh. 27 - Prob. 58APCh. 27 - Prob. 59APCh. 27 - Prob. 60APCh. 27 - Prob. 61APCh. 27 - Prob. 62APCh. 27 - Prob. 63APCh. 27 - Review. An office worker uses an immersion heater...Ch. 27 - Prob. 65APCh. 27 - Prob. 66APCh. 27 - Prob. 67APCh. 27 - Prob. 68APCh. 27 - Prob. 69APCh. 27 - Prob. 70APCh. 27 - Prob. 71APCh. 27 - Prob. 72APCh. 27 - Prob. 73APCh. 27 - Prob. 74APCh. 27 - Prob. 75APCh. 27 - Prob. 76APCh. 27 - Review. A parallel-plate capacitor consists of...Ch. 27 - The dielectric material between the plates of a...Ch. 27 - Prob. 79APCh. 27 - Prob. 80APCh. 27 - Prob. 81APCh. 27 - Prob. 82CPCh. 27 - Prob. 83CPCh. 27 - Material with uniform resistivity is formed into...Ch. 27 - Prob. 85CP
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