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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 (20.0โˆ˜C)the ammeter reads 18.5Awhile at 92.0โˆ˜Cit

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 at20โˆ˜Cis1.06ร—10-5ฮฉ.m.

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

Step by step solution

01

Define the ohmโ€™s law, resistance(R)and resistivity(ฯ).

Write the expression for the Ohmโ€™s law.

V=IR

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

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

R=VI=ฯLA

Here,ฯis resistivityฮฉ.m, L is length in m and A is area inm2.

Writeฯof the material is the ratio of the magnitude of the electric fieldEand

current density J .

ฯ=EJฯT=ฯ01+ฮฑT-T0

Write the expression for the temperature coefficient as:

ฮฑ=RTR0-1T-T0

02

Determine the resistivityฯ.

(a)

Consider the given parameters:

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

The resistivity of cylindrical rod is:

ฯ=R0AL

Substitute the value and solve as,

ฯ=15.018.50ฯ€0.500221.50โˆตR0=VI0=0.811ฯ€0.00251.50=1.60ร—10-5ฮฉ.m

Hence, the resistivity at 20โˆ˜Cis 1.60ร—10-5ฮฉ.m.

03

Determine the temperature coefficient ฮฑ.

(b)

Consider the given parameters:

T0=20.0โˆ˜CT=92.0โˆ˜CI=17.2A

The temperature coefficient ฮฑis:

ฮฑ=RTR0-1T-T0

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

Substitute the value and solve.

ฮฑ=15/17.20.811-192-20=0.872/0.811-172=0.00105Cโˆ˜-1

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

0.00105Cโˆ˜-1.

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