Question

One junction of a certain thermoelectric couple is at a fixed temperature T_r and the other junction is at temperature T. The thermo electromotive force for this is expressed by $E = \mspace{6mu} K(T - \mspace{6mu} T_{r})\mspace{6mu}\left\lbrack T_{0}\mspace{6mu} - \mspace{6mu}\frac{1}{\mspace{6mu} 2\mspace{6mu}}\mspace{6mu}(T\mspace{6mu} + \mspace{6mu} T_{r}) \right\rbrack$at temperature

T = $\frac{1}{2}\mspace{6mu} T_{0}$, the thermo electric power will be

• A. $\frac{1}{2}\mspace{6mu} KT_{0}$
• B. $KT_{0}$
• C. $\frac{1}{2}\mspace{6mu} KT_{0}^{2}$
• D. $\frac{1}{2}\mspace{6mu} K(T_{0}\mspace{6mu} - \mspace{6mu} T_{r})^{2}$

Answer 1

Answer: (A)

$\frac{1}{2}\mspace{6mu} KT_{0}$

Answer 2

As we know thermo electric power $S = \frac{dE}{dT}$. Hence by differentiating the given equation and putting $T = \frac{1}{2}\mspace{6mu} T_{0}$ we get $S = \frac{1}{2}KT_{0}$.