Respuesta :
Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =[tex]R_T =\frac{\rho_T \ l}{A}[/tex]
Where [tex]R_T[/tex] =Resistance of wire at Temperature T
[tex]\rho_T[/tex] = Resistivity at temperature T [tex]=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )][/tex]
Where [tex]T_0 =20\ Deg\ C , \ \rho_0 = Constant, \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)[/tex]
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows
[tex]\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}[/tex]
[tex]\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}[/tex]
[tex]1.24=1+\alpha (T-20)[/tex]
[tex]0.24=\alpha(\ T -20 )[/tex]
[tex]Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1[/tex]
T = 81.52 Deg C
At the temperature of [tex]81.53^{0} C[/tex], resistance will be [tex]0.31[/tex] ohm.
Resistance :
The resistance at different temperature is given as,
[tex]R_{2}=R_{1}(1+ \alpha(T_{1}-T_{0}))[/tex]
- [tex]\alpha[/tex] is the temperature coefficient of resistivity
- [tex]T_{0}[/tex] is a reference temperature
- [tex]R_{2}[/tex] is resistance at temperature [tex]T_{1}[/tex]
Given that, [tex]R_{2}=0.31,R_{1}=0.25,T_{0}=20,\alpha=3.9*10^{-3}[/tex]
Substitute values in above relation.
[tex]0.31=0.25(1+3.9*10^{-3}(T_{1}-20) )\\\\\frac{0.24}{3.9*10^{-3} } =(T_{1}-20)\\\\T_{1}=61.53+20=81.53^{0} C[/tex]
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