Resistance and Resistivity of tungsten wire

[Unknowned]

please check my work someone.

A tungsten wire has a radius of .075mm and is heated from 20.0 to 1320 degrees C. The temperature coefficient of resistivity is 4.5x10^-3 (C)^-1. When 120V is applied across the ends of the hot wire, a current os 1.5A is produced. How long is the wire? Neglect anf effects due to thermal expansion.

OKay we all know R= p x L/A where R is the resistance, p is the proportionality constant known as the resistivity of the material, L is the length, and A is the area.

the book says 5.6x10^-8 is the resistivity for a tungsten wire.

So we solve the equation for L, L= R(A)/p

R= V/I soo 120v/1.5A= 80ohms R= 80ohms

A= 4x3.14xradius^2 so 4x3.14x(.075mm)^2= .071mm^2 A= .071mm^2 or 7.1x10^-5m

T= Tc + 273 20c+(273c)=293K and 1320c+(273c)=1593.15K 1593.15k-293k=1300.12 K

P= (Resistivity)[1+(coefficiant of resistivity)(1300.12K)
p= (5.6x10^-8)[1+(4.5x10^-3)(1300.12)= 3.84x10^-7 p=3.83x10^-7

L=80A(7.1x10^-5m)/(3.83x10^-7ohmsxmeters)=14830m long

is that correct?? anyone??

 

[mikey23]

A= Pi x radius^2 so 3.14x(.075mm)^2= 17.67x10^-9 m^2

T= 1320-20=1300 C

P= (Resistivity)[1+(coefficiant of resistivity)(T)]
p= (56x10^-9)[1+(4.5x10^-3)(1300)] = 383.6x10^-9

L=80(17.67x10^-9)/(383.6x10^-9)=3.685m long

This seems intuitively long to me, so if the resistivity figure is ohm-cm then this seems OK at 3.685cm.

On the other hand, 1320 C is not even a glowing red, is it? So this is not a 180W light bulb, so 3.685m may be OK.

I wonder what it would take to get it up to 3000 - 4000 C like an ordinary light bulb? I guess I will save that for another rainy day. https://www.physicsforums.com/Nexus/editor/smilie.png