Q. If the emf of a thermocouple, one junction of which is kept $0℃$ is given by $e = at + \frac{1}{2}\ bt^{2},$ then the neutral temperature will be
A) $\frac{a}{b}$
B) $- \frac{a}{b}$
C) $\frac{a}{2b}$
D) $- \frac{1}{ab}$
Explanation
(b) When temperature of the hot junction of a thermocouple rises, the thermo-emf increases and becomes maximum at a particular temperature. This temperature is called neutral temperature.
Given, $e = at + \frac{1}{2}\ bt^{2}$
Differentiating with respect to $t,$ we get
$\frac{de}{dt} = a + bt$
For maximum value of $e,$
$\frac{de}{dt} = 0$
$\therefore a + bt = 0$
$\Rightarrow t = - \frac{a}{b}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.