$e = \alpha\ t - \frac{1}{2}\beta\ t^{2},$ if temperature of cold junction is $0℃$ then temperature of inversion is (if $\alpha = 500.0\mu V/℃,\beta = 5.0\mu V/square℃$)

Question

Dr. Abubacker · Accepted Answer

(b) $t_{n} = \frac{\alpha}{\beta} = \left( \frac{500}{5} \right) = 100℃$
Also$\ t_{n} = \frac{t_{i} + t_{c}}{2} \Rightarrow 100 = \frac{t_{i} + 0}{2} \Rightarrow t_{i} = 200℃$

Q. $e = \alpha\ t - \frac{1}{2}\beta\ t^{2},$ if temperature of cold junction is $0℃$ then temperature of inversion is (if $\alpha = 500.0\mu V/℃,\beta = 5.0\mu V/square℃$)

Explanation

Reviewed by Dr. Abubacker

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