The rate of increase of thermo-e.m.f. with te | vedclass

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(B) The thermo-e.m.f. $(E)$ in a thermocouple is given by the relation $E = \alpha 	heta + \frac{1}{2} \beta 	heta^2$,where $	heta$ is the temperat

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is zero

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(B) The thermo-e.m.f. $(E)$ in a thermocouple is given by the relation $E = \alpha 	heta + \frac{1}{2} \beta 	heta^2$,where $	heta$ is the temperature difference between the junctions.At the neutral temperature $(	heta_n)$,the thermo-e.m.f. is maximum.Mathematically,for the e.m.f. to be maximum,the first derivative of $E$ with respect to temperature must be zero.Therefore,$\frac{dE}{d	heta} = 0$ at $	heta = 	heta_n$.This derivative $\frac{dE}{d	heta}$ represents the rate of increase of thermo-e.m.f. with temperature.Thus,at the neutral temperature,the rate of increase of thermo-e.m.f. is zero.

The rate of increase of thermo-e.m.f. with temperature at the neutral temperature of a thermocouple [$AIPMT$ $2011$]

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