Question

A metal wire has a resistance of 14.00 Ω at a temperature of 25.0°C. If the same wire has a resistance of 14.55 Ω at 90.0°C, what is the resistance when its temperature is −32.0°C?

Answer 1

Answer:

13.52 Ω

Explanation:

coefficient of thermal resistance be α

R₀ , R₂₅ , R₉₀ and R₋₃₂ be resistances at 0 , 25 , 90 , and - 32 degree

R₂₅ = R₀ + α x 25

R₉₀ = R₀ + α x 90

R₉₀ - R₂₅ = 65 x α

α = (R₉₀ - R₂₅ )/ 65

= (14.55 - 14) / 65

=   .55 / 65 Ω per °C,

R₂₅ = R₀ + α x 25

14 = R₀ + (.55 / 65 )x 25

=  R₀ + .2115

R₀ = 13.7885 Ω

R₋₃₂ = R₀ - α x 32

= 13.7885 -(  .55 / 65) x 32

=  13.7885 - .27077

= 13.51773 Ω

= 13.52 Ω

Answer 2

Answer:

Explanation:

resistance at 25°C, R' = 14 ohm

resistance at 90°C, R'' = 14.55 ohm

Let R be the resistance at - 32°C. Let Ro be the resistance at 0°C. Let α be the temperature coefficient of resistance.

R = Ro ( 1 + αΔT)

14 = Ro ( 1 + α x 25) .... (1)

14.55 = Ro( 1 + α x 90) .... (2)

Divide equation (2) by equation (1)

$\frac{14.55}{14}=\frac{1+90\alpha }{1+25\alpha }$

14.55 + 363.75 α = 14 + 1260 α

896.25 α = 0.55

α = 6.14 x 10^-4 / °C

So,

R = Ro ( 1 + 32 α) .... (3)

Divide equation (3) by equation (1)

$\frac{R}{14}=\frac{1+32\alpha }{1+25\alpha }$

$R=14\times \frac{1+32\times 6.14\times 10^{-4}}{1+25\times 6.14\times 10^{-4}}$

R = 14.06 ohm