Question

The thermo-emf of a thermocouple is $25\mu V{/}^{o}C$ 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 the system is

• A. ${12}^{o}C$
• B. $8^{o}C$
• C. ${20}^{o}C$
• D. ${16}^{o}C$

Answer 1

The correct option is D ${16}^{o}C$

$△\theta =\frac{40ohm\times {10}^{-5}A}{25\times {10}^{-6}\frac{\mu V}{{}^{o}C}}={16}^{o}C$