Acylindrical 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?

express your answer using two significant figures.

Explanation:

R = ρ L / S

ρ is resistivity , L is length of wire and S is cross sectional area

R at 20 degree

=

R₁ = 78.95 x 10⁻⁴ Ω

Resistivity at 120 degree

= 5.25 x 10⁻⁸ ( 1 + 5.25 x 10⁻⁸ x .0045 x 100)

= 7.61 x 10⁻⁸

Resistance at 120 degree

=

R₂= 114.44 x 10⁻⁴ Ω

Potential difference at 20 degree

= current x resistance

= 12.5 x 78.95 x 10⁻⁴

= 986.875 x 10⁻⁴ V

Electric field at 20 degree

= Potential diff / length

= 986.875 x 10⁻⁴ / 17 x 10⁻²

= 58 x 10⁻² V/m

Potential difference at 120 degree

= current x resistance

= 12.5 x 114.44 x 10⁻⁴

= 1430.5  x 10⁻⁴ V

Electric field at 120 degree

= Potential diff / length

= 1430.5 x 10⁻⁴ / 17 x 10⁻²

= 84.15  x 10⁻² V/m

A ) Maximum electric field = 84.15  x 10⁻² V/m

B )  114.44 x 10⁻⁴  Ω

C ) 1430.5  x 10⁻⁴ V

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000