Question

A cylindrical tungsten filament 16.0 cm long with a diameter of 1.00 mm 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 13.5 A at all temperatures ( Resistivity of tungsten at 20 ∘C is 5.25×10−8Ω⋅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?
Express your answer using two significant figures.

Answer 1

(a) 1.3 V/m

(b)
`R\approx 1.6*10^(-2)` Ω

(c) 0.22 V

Step-by-step explanation:

From Ohms law

`J=\sigma E` or
`J=\frac {E}{\rho}` or
`\frac {I}{A}=\frac {E}{\rho}` hence
`E=\rho \frac {I}{A}`

At
`120^(\circ)`

`\rho=5.25*10^(-8)+5.25*10^(-8)(0.0045)*(120-20)=7.61*10^(-8)` Ω⋅m

Area of cylinder is given by

`A=\pi (\frac {d}{2})^(2)=\pi (\frac {1*10^(-3)}{2})^(2)=7.85398*10^(-7)`

Maximum electric field is given by

`E=\frac {I\rho}{A}=\frac {13.5*(7.6125*10^(-8))}{7.85398*10^(-7)}=1.308492365 V/m`

`E\approx 1.3 V/m`

(b)

Resistance of electric field

`R=\frac {\rho_(120)L}{A}=\frac {7.6125*10^(-8)*(16*10^(-2))}{7.85398*10^(-7)}=0.015508058`

`R\approx 1.6*10^(-2)` Ω

(c

Maximum potential drop

`\triangle V=IR=13.5*0.016=0.216 V`

`\triangle V\approx 0.22 V`