Question

The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where T0 is a reference temperature, usually 20°C, and α is the temperature coefficient of resistivity. For copper, α = 3.9× 10−3 ° C−1. Suppose a long, thin copper wire has a resistance of 0.25 Ω at 20°C.
At what temperature, in °C, will its resistance be 0.31 Ω?

Answer 1

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire =$R_T =\frac{\rho_T \ l}{A}$

Where $R_T$ =Resistance of wire at Temperature T

$\rho_T$ = Resistivity at temperature T $=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]$

Where $T_0 =20\ Deg\ C , \ \rho_0 = Constant, \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)$

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

$\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}$

$\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}$

$1.24=1+\alpha (T-20)$

$0.24=\alpha(\ T -20 )$

$Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1$

T = 81.52 Deg C