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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??
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??