The temperature at which the resistance is [tex]1 k\Omega[/tex] is [tex]89.4^{\circ}C[/tex]
Explanation:
For the thermistor in this problem, the relationship between temperature and resistance is linear.
We have:
[tex]R_1 = 5000 \Omega[/tex] when the temperature is [tex]T=25^{\circ}C[/tex]
[tex]R_2=340 \Omega[/tex] when the temperature is [tex]T=100^{\circ}C[/tex]
Assuming a straight-line relationship, we can find the slope of the line:
[tex]m=\frac{R_2-R_1}{T_2-T_1}=\frac{340-5000}{100-25}=-62.1 \Omega/^{\circ}C[/tex]
Now that we know the slope, we can extrapolate the temperature when the resistance is
[tex]R_3 = 1 k\Omega = 1000 \Omega[/tex]
In fact, we can write:
[tex]m=\frac{R_3-R_2}{T_3-T_2}[/tex]
And solving for [tex]T_3[/tex],
[tex]m(T_3-T_2)=R_3-R_2\\T_3 = T_2 +\frac{R_3-R_2}{m}=100 + \frac{1000-340}{-62.1}=89.4^{\circ}C[/tex]
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