The thermoelectric electromotive force e is g | vedclass

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(B) The thermoelectric $EMF$ is given by $e = \alpha t - \frac{1}{2}\beta t^2$.For the neutral temperature $t_n$,the $EMF$ is maximum,so $\frac{de}{dt

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$200 \, ^\circ C$

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(B) The thermoelectric $EMF$ is given by $e = \alpha t - \frac{1}{2}\beta t^2$.For the neutral temperature $t_n$,the $EMF$ is maximum,so $\frac{de}{dt} = 0$.$\frac{de}{dt} = \alpha - \beta t = 0 \Rightarrow t_n = \frac{\alpha}{\beta}$.Substituting the given values: $t_n = \frac{500}{5} = 100 \, ^\circ C$.The relationship between neutral temperature $t_n$,inversion temperature $t_i$,and cold junction temperature $t_c$ is $t_n = \frac{t_i + t_c}{2}$.Given $t_c = 0 \, ^\circ C$,we have $100 = \frac{t_i + 0}{2}$.Therefore,$t_i = 200 \, ^\circ C$.

The thermoelectric electromotive force $e$ is given by $e = \alpha t - \frac{1}{2}\beta t^2$. If the temperature of the cold junction is $0 \, ^\circ C$,find the temperature of inversion $t_i$. (Given: $\alpha = 500.0 \, \mu V/^\circ C$,$\beta = 5.0 \, \mu V/^\circ C^2$)

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