A 1.30m cylindrical rod of diameter 0.550cm is connected to a power supply that

Question

A 1.30m cylindrical rod of diameter 0.550cm is connected to a power supply that maintains a constant potential difference of 15.0V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 degree C)the ammeter reads 18.3 A, while at 92.0 degree C it reads 17.3A. You can ignore any thermal expansion of the rod. Find the resistivity and for the material of the rod at 20 degree C. Part B Find the temperature coefficient of resistivity at 20 degree C for the material of the rod.

Explanation / Answer

Here ,

area = pi*r^2

area = 3.141*0.00275^2

area = 2.37 *10^-5 m^2

Now ,

R = p*L/A

Using ohm's law

15/18.3 = p*1.30/2.37 *10^-5

p = 1.498 *10^-5 Ohm.m

the resistivity of the material at 20 C is 1.498 *10^-5 Ohm.m

B)

at T = 20 C

R1 = 15/18.3

R1 = 0.82 Ohm

at T = 92 C

R2 = 15/17.3

R2 = 0.867 Ohm

here ,

R2 = R1 *(1+a*del T )

0.867 = 0.82*(1 + a*(92 - 20))

a = 7.97 *10^-4 oC-1

the temperature coeffiecent of resistivity is 7.97 *10^-4 oC-1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.