Q. A thermocouple develops $40\ \mu V/kelvin.$ If hot and cold junctions are at $40℃$ and $20℃$ respectively, then then emf developed by a thermopile using such 150 thermocouples in series shall be
A) $150mV$
B) $80mV$
C) $144mV$
D) $120mV$
Explanation
(d) The temperature difference is $20℃ = 20\ K$. So that thermo $e$mf developed $E = \alpha\theta = 40\frac{\mu V}{K} \times 20K = 800\ \mu V$
Hence total $e$mf $= 150 \times 800 = 12 \times 10^{4}\ \mu V = 120\ mV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.