A1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 15.0 v across its ends, while an ammeter measures the current through it. you observe that at room temperature (20.0°c) the ammeter reads 18.6 a, while at 92.0°c it reads 17.5 a. you can ignore any thermal expansion of the rod. (a) find the resistivity at 20°c for the material of the rod. (b) find the temperature coefficient of resistivity at 20°c for the material of the rod.

(a)

The resistance of the rod is given by:

(1)

where

is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: , so the cross-sectional area is

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

And now we can re-arrange the eq.(1) to solve for the resistivity:

(b)

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

The equation that gives the change in resistance as a function of the temperature is

where

is the resistance at the new temperature (92.0°C)

is the resistance at the original temperature (20.0°C)

is the temperature coefficient of resistivity

Solving the formula for , we find

6/12=1/2

step-by-step explanation:

the amount of current in the coil