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A tungsten wire has a radius of 0.073 mm and is heated from 20.0 to 1500 °C. The temperature coefficient of resistivity is α = 4.5 10-3 (C°)−1. When 110 V is applied across the ends of the hot wire, a current of 1.6 A is produced. How long is the wire? Neglect any effects due to thermal expansion of the wire.

physics
 Apr 29, 2016

Best Answer 

 #1
avatar+33603 
+10

Resistance of wire at 1500°C is  R = 110V/1.6A  or  R = 68.75 ohms 

 

Electrical resistivity of tungsten at 20°C is  rho20 = 5.5*10-8 ohm.m 

 

Electrical resistivity of tungsten at 1500°C is  rho1500 = rho20*(1 + alpha*deltaT)

                                                                       rho1500 = 5.5*10-8*(1 + 4.5*10-3*(1500-20)) = 4.213*10-7 ohm.m

 

Length of tungsten wire at 1500°C  L = rho1500/R  or  L = 4.213*10-7 ohm.m/68.75ohms  ≈  6*10-9 m

 

 

Hmm!  Seems very short!  Someone needs to check this.

 Apr 30, 2016
 #1
avatar+33603 
+10
Best Answer

Resistance of wire at 1500°C is  R = 110V/1.6A  or  R = 68.75 ohms 

 

Electrical resistivity of tungsten at 20°C is  rho20 = 5.5*10-8 ohm.m 

 

Electrical resistivity of tungsten at 1500°C is  rho1500 = rho20*(1 + alpha*deltaT)

                                                                       rho1500 = 5.5*10-8*(1 + 4.5*10-3*(1500-20)) = 4.213*10-7 ohm.m

 

Length of tungsten wire at 1500°C  L = rho1500/R  or  L = 4.213*10-7 ohm.m/68.75ohms  ≈  6*10-9 m

 

 

Hmm!  Seems very short!  Someone needs to check this.

Alan Apr 30, 2016
 #2
avatar+1036 
+10

Solution:

 

 

\(R = \dfrac{110V}{1.6A} = 68.75 {~ohms} \\ \rho_{20} = 55.0*10^{-9} \Omega \cdot M \\ \alpha = \text {Temperature coefficient for tungsten} = 0.0045K^{-1} (K^{-1} \equiv C^{-1})\\ \rho_{1500}= (\rho_{20})*[1+(\alpha\Delta T)]\\ \rho_{1500}= (55.0*10^{-9})*(1+0.0045 C^{-1})*(1500-20)C =8.287125*10^{-5} \text {M} \cdot \Omega \\ A_c = (\pi * r^2) = \pi * (0.073*10^{-6})^2 = 1.67415473*10^{-8} M^2 \text {(Area cross-section)} \\\\ \)

 

\(R= \rho *(L*A) \rightarrow \text {Solve for length L. } \\ L = \dfrac {RA} {(\rho_{20}*[1 + \alpha * \Delta T])} \\ \hspace {10 mm} \\ \dfrac {(68.75)*(16.7415473*10^{-9})} {(55.0*10^{-9})*(1 + [0.0045 *1480])} =2.73198 ~\text {Meters} \\ \)

 May 1, 2016

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