Q. The $emf$ of a thermocouple, one junction of which is kept at $0℃,$ is given by $e = at + bt^{2}.$ The Peltier co-efficient will be
A) $(t + 273)(a + 2bt)$
B) $(t + 273)(a - 2bt)$
C) $(t - 273)(a - 2bt)$
D) $(t - 273)(a + 2bt)$
Explanation
(a) $\because$ Peltier coefficient$\ \pi = T\frac{de}{dT}\ $and $t℃ = T - 273$
$\therefore e = a(T - 273) + b(T - 273)^{2}$
Differentiating $w.r.t.\ T\frac{de}{dT} = a + 2b(T - 273)$
$\pi = T\frac{de}{dT} = T\left\lbrack a + 2b(T - 273) \right\rbrack \Rightarrow \pi = (t + 273)(a + 2bt)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.