Q. One junction of a certain thermo-couple is at a fixed temperature $T_{r}$ and the other junction is at temperature $T$. The thermo electric force for this is expressed by $E = K\left( T - T_{r} \right)\lbrack\ T_{0} + \frac{1}{2}\left( T^{2} + T_{r}^{2})\ \right\rbrack$. At temperature $T = T_{0}/2$ the thermoelectric power is
A) $\frac{1}{2}\ K\ T_{0}$
B) $\frac{3}{2}$ $K\ T_{0}$
C) $\frac{1}{2\ }\ K\ T_{0}^{2}$
D) $\frac{1}{2\ }K\ \left( T_{0} - T_{r} \right)^{2}$
Explanation
(b) $E = K\left( T - T_{r} \right)T_{0} + \frac{1}{2}K(T^{2} - T_{r}^{2})$
$\frac{dE}{dT} = KT_{0} + \frac{1}{2}K \times 2T = KT_{0} + KT$
At temperature $T = T_{0}$/2,
Thermo-electric power is $\frac{dE}{dT} = KT_{0} + K\frac{T_{0}}{2} = \frac{3}{2}K{\ T}_{0}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.