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A 1.50-mcylindrical rod of diameter 0.500cmis connected to

a power supply that maintains a constant potential difference of 15.0Vacross

its ends, while an ammeter measures the current through it. You observe that

at room temperature (20.0C)the ammeter reads 18.5Awhile at 92.0Cit

reads 17.2A. You can ignore any thermal expansion of the rod. Find (a) the

resistivity at and (b) the temperature coefficient of resistivity at for the material of the rod.

Short Answer

Expert verified

(a)The resistivity at20Cis1.06×10-5Ω.m.

(b)The temperature coefficient of resistivity at for the material of the rod is0.00105C-1.

Step by step solution

01

Define the ohm’s law, resistance(R)and resistivity(ρ).

Write the expression for the Ohm’s law.

V=IR

Here, I is current in ampereA,Ris resistance in ohmsΩandVis the potential difference voltV.

Write the expression for the resistance in terms of the voltage and resistivity.

R=VI=ρLA

Here,ρis resistivityΩ.m, L is length in m and A is area inm2.

Writeρof the material is the ratio of the magnitude of the electric fieldEand

current density J .

ρ=EJρT=ρ01+αT-T0

Write the expression for the temperature coefficient as:

α=RTR0-1T-T0

02

Determine the resistivityρ.

(a)

Consider the given parameters:

L=1.50md=0.500cmV=15VI0=18.50A

The resistivity of cylindrical rod is:

ρ=R0AL

Substitute the value and solve as,

ρ=15.018.50π0.500221.50R0=VI0=0.811π0.00251.50=1.60×10-5Ω.m

Hence, the resistivity at 20Cis 1.60×10-5Ω.m.

03

Determine the temperature coefficient α.

(b)

Consider the given parameters:

T0=20.0CT=92.0CI=17.2A

The temperature coefficient αis:

α=RTR0-1T-T0

Here, RT=VIand R0is 0.811 role="math" localid="1655788149360" Ω.

Substitute the value and solve.

α=15/17.20.811-192-20=0.872/0.811-172=0.00105C-1

Hence, temperature coefficient of resistivity at 20Cfor the material of the rod is

0.00105C-1.

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