Question

A cylindrical tungsten filament 17.0cm long with a diameter of 1.20mm is to be used in a machine for which the temperature will range from room temperature (20 ?C) up to 120 ?C. It will carry a current of 12.5A at all temperatures ( Resistivity of tungsten at 20 ?C is 5.25�10?8??m , the temperature coefficient of resistivity at 20 ?C is 0.0045 ?C?1 )

A) What will be the maximum electric field in this filament?

Express your answer using two significant figures.

B) What will be its resistance with that field?

Express your answer using two significant figures.

C) What will be the maximum potential drop over the full length of the filament?

Answer 1

We need to use the Ohms law to give a proper solution,

`J=\sigma E\\J= (E)/(\rho)\\(I)/(A)=(E)/(\rho)\\E=(I)/(A)\rho`

The temperature is 120° c, so calculate
`\rho`

`\rho = 5.25*10^(-8)+5.25*10^(-8)(0.0045)(120-20)`

`\rho = 7.61*10^(-8) \Omega .m`

We can get the Are of the cylinder,

`A= \pi (d^2)/(4)\\A= \pi (0.6*10^(-3))^2\\A=1.13*10^(-6)m^2`

To get the electric field replace the values,

`E=(\rho I)/(A)\\E= (7.61*10^(-8)12.5)/(1.13*10^(-6))\\E=0.84N/c`

`Solving for the resistance, we haveR=(\rho L)/(A)\\R= (7.61*10^(-8)0.17)/(1.13*10^-6)\\R=0.01\Omega`

At the end we can get the Potential

`V_(max)=IR\\V_(max)=(12.5)(0.01)\\V_(max)=0.12V`