A 1.90-m cylindrical rod of dimeter 0 540 cm is connected to a power supply that

Question

A 1.90-m cylindrical rod of dimeter 0 540 cm is connected to a power supply that maintains constant potential difference of 15.0 V across is ends while an ammeter be 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 17.4 K. You can ignore any thermal of the rod (a) Find the resistivity at 20 degree C for the of the rod the tempera resistivity at 20degreeC for the material of the rod Ans 9.88c06 ohm in. 0.000118 per degree Celsius.

Explanation / Answer

In my college physics book, I found the following equations.
Resistance = * (L/A) and Rf = Ri * ([1 + * (Tf – Ti)]

= Resistivity
L = length in meters
A = cross sectional area in m^2
= temperature coefficient of resistivity

L = 1.90 m
Area = * r^2
r = d/2 = 0.27 cm = 2.7 x 10^-3 m
Area = * (2.7 x 10^-3)^2 = 2.289 x 10^-5 m^2


The cylindrical rod is similar to a resistor. Since the current is decreasing, the resistance must be increasing. This means the resistance is increasing as the temperature increases.
Resistance = Voltage ÷ Current
At 20, R- = 15/18.3 = 0.8197 ohm
At 92, Rf = 15/17.4 = 0.862 ohm


Now you know the resistance at the two temperatures. Let’s determine the resistivity at the two temperatures.
Resistance = * (L/A)
= Resistance * (A/L)

At 20, = (0.8197 x 2.289 x 10^-5) / 1.90 = 0.9875 x 10^-5 ohm.meter
At 92, = (0.862 x 2.289 x 10^-5) / 1.90 = 1.038 x 10^-5 ohm.meter

Now you know the resistivity at the two temperatures. Let’s determine the temperature coefficient of resistivity for the material of the rod.

Rf = Ri * ([1 + * (Tf – Ti)]
0.862 = 0.8197( 1 + *72)

0.862 = 0.8197 + 59

= 0.000716949 = 7.169 x 10^-4 per degree celcius

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