Respuesta :
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:
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)
[tex]\frac{14.55}{14}=\frac{1+90\alpha }{1+25\alpha }[/tex]
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)
[tex]\frac{R}{14}=\frac{1+32\alpha }{1+25\alpha }[/tex]
[tex]R=14\times \frac{1+32\times 6.14\times 10^{-4}}{1+25\times 6.14\times 10^{-4}}[/tex]
R = 14.06 ohm
