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The rate of increase of thermo-emf with temperature at the neutral temperature of a thermocouple
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1456 Upvotes
Verified Answer
The correct answer is:
is zero
We have
$$
e=a t+b t^2
$$
Natural temperature $\frac{d e}{d t}=a+2 b t$
$$
T_n=-\frac{a}{2 b}
$$
At netural temperature
$$
\frac{d e}{d t}=0
$$
$$
e=a t+b t^2
$$
Natural temperature $\frac{d e}{d t}=a+2 b t$
$$
T_n=-\frac{a}{2 b}
$$
At netural temperature
$$
\frac{d e}{d t}=0
$$
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