A platinum resistance thermometer measure temperature on the absorption that the change in electrical resistance of platinum is proportional to the change in temperature. The resistance of the platinum wire at zero degrees Celsius is 10.4ohms and 10.71ohms at 100°c
(1) what will be the resistance of the wire when the temperature is 45°c
Rp(0°C)=10.4 ΩRp(100°C)=10.71 ΩRp(45°C)= ?R_{p(0°C)} = 10.4 \ \Omega\\ R_{p(100°C)} = 10.71 \ \Omega\\ R_{p(45°C)} = \ ?Rp(0°C)=10.4 ΩRp(100°C)=10.71 ΩRp(45°C)= ?
100°C−0°C45°C−0°C=10.71−10.410.71−Rp(45°C)\dfrac{100°C - 0°C}{45°C - 0°C} = \dfrac{10.71-10.4}{10.71 - R_{p(45°C)}}45°C−0°C100°C−0°C=10.71−Rp(45°C)10.71−10.4
10045=0.3110.71−Rp(45°C)\dfrac{100}{45} = \dfrac{0.31}{10.71 - R_{p(45°C)}}45100=10.71−Rp(45°C)0.31
209=0.3110.71−Rp(45°C)\dfrac{20}{9} = \dfrac{0.31}{10.71 -R_{p(45°C)}}920=10.71−Rp(45°C)0.31
20(10.71−Rp(45°C))=0.31×920(10.71-R_{p(45°C)}) = {0.31× 9}20(10.71−Rp(45°C))=0.31×9
214.2−20Rp(45°C)=2.79214.2 - 20R_{p(45°C)} =2.79214.2−20Rp(45°C)=2.79
Rp(45°C)=214.2−2.7920R_{p(45°C)} = \dfrac{214.2 -2.79}{20}Rp(45°C)=20214.2−2.79
Rp(45°C)=10.571 ΩR_{p(45°C)} = 10.571 \ \OmegaRp(45°C)=10.571 Ω
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I really love your hard work. It's helping me alot! God bless you!