Question

The e.m.f. developed in a thermocouple is given by $E=\alpha t+\frac{1}{2}\beta t^2$ where t is the temperature of hot junction, cold junction being at $0^0$C. The thermoelectric power of this thermocouple is

• A. $\alpha +\frac{\beta t}{2}$
• B. $\alpha +\beta t$
• C. $\frac{\alpha +t^2}{2}$
• D. $\frac{\alpha }{2\beta }$

Answer 1

The correct option is B $\alpha +\beta t$

Thermoelectric power( $\text{Seebeck coefficient}$ ) of a material is the measure of the magnitude of an induced thermoelectric voltage in response to a temperature difference across that material, as induced by the Seebeck effect.

Given:

$E$ = $\alpha t$ $+$ $\frac{1}{2}$$\beta$$t^2$

Seebeck coefficient($S$) = $\frac{\partial E}{\partial T}$ = $\alpha$ $+$ $\beta t$

Thermoelectric power of this thermocouple = $\frac{\partial E}{\partial T}$ ${}_{(t=t)}$

= $\alpha$ $+$ $\beta t$

Therefore, thermoelectric power of this thermocouple is $\alpha$ $+$ $\beta t$.

Hence, the correct option is ${}^{\prime }{B}^{\prime }$.