Q. The thermo-emf of a thermocouple is $25\mu\ V/℃$ at room temperature. A galvanometer of $40\ \Omega$ resistance, capable of detecting current as low as $10^{- 5}\ A,$ is connected with the thermocouple. The smallest temperature difference that can be detected by this system is
A) $16℃$
B) $12℃$
C) $8℃$
D) $20℃$
Explanation
(a) Thermo-emf of thermocouple $= 25\frac{\mu V}{℃}.$
Let $\theta$ be the smallest temperature difference.
Therefore, after connecting the thermocouple with the galvanometer, thermo-emf
$E = \left( 25\frac{\mu V}{℃} \right) \times \theta(℃)$
$\ \ \ \ = 25\theta \times 10^{- 6}\ V$
Potential drop developed across the galvanometer
$= iR = 10^{- 5} \times 40 = 4 \times 10^{- 4}\ V$
$\therefore 4 \times 10^{- 4} = 25\theta \times 10^{- 6}$
$\therefore\theta = \frac{4}{25} \times 10^{2} = 16℃\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.