Q. One junction of a thermo-couple is a particular temperature $T_{r}$ and another is at $T$. Its thermo emf is expressed as $E = K(T - T_{r})\left\{ T_{0} - \frac{1}{2}(T + T_{r}) \right\}$ At a temperature $T = \frac{T_{0}}{2}$, the value of thermo-electric power will be
A) $\frac{1}{2}KT_{0}$
B) $KT_{0}$
C) $\frac{1}{2}KT_{0}^{2}$
D) $\frac{1}{2}K\left( T_{0} - T_{r} \right)^{2}$.
Explanation
(a) Thermo electric power $E = k\left( TT_{0} - T_{0}T_{r} - \frac{1}{2}T_{0}^{2} + \frac{1}{2}T_{r}^{2} \right)$
$\ S = \frac{dE}{dT} = k\left\lbrack T - 0 - \frac{1}{2} \times 2T + 0 \right\rbrack$
$= k\lbrack T_{0} - T\rbrack$
At $\ \ T = T_{0}/2$
$\ S = \frac{1}{2}kT_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.