Problem $2\mathrm{.}3$

Thermistors are temperature$-$measuring devices based on the principle that the thermistor material exhibits a

change in electrical resistance with a change in temperature. By measuring the resistance of the thermistor

material, one can then determine the temperature. For a $10$K$3$A Betatherm thermistor, the relationship between

Figure $1$: A typical thermistor

the resistance $R$ of the thermistor and the temperature $T$ is given by

$\frac{1}{T}=1\mathrm{.}129241{10}^{-3}+2\mathrm{.}341077{10}^{-4}logR+8\mathrm{.}775468**{10}^{-8}(logR{)}^3$

where $T$ is in Kelvin and $R$ is in Ohms, and log denotes the natural logarithm. A thermistor error of no more

than $+-0\mathrm{.}01C$ is acceptable. To find the range of the resistance that is within this acceptable limit at $19C,$ we

need to solve

$\frac{1}{19\mathrm{.}01+273\mathrm{.}15}=1\mathrm{.}129241{10}^{-3}+2\mathrm{.}341077{10}^{-4}logR+8\mathrm{.}775468**{10}^{-8}(logR{)}^3$

$\frac{1}{18\mathrm{.}99+273\mathrm{.}15}=1\mathrm{.}129241{10}^{-3}+2\mathrm{.}341077{10}^{-4}logR+8\mathrm{.}775468**{10}^{-8}(logR{)}^3$

Write a computer routine implementing Newton's method and solve equations $(1)$ and $(2)$ using Newton's

method with initial guess $R_0=15000$ and error tolerance of ${10}^{-6}.$

What is the obtained range for resistance values?