Question

The thermo e.m.f. of a thermocouple is $25\mu V{/}^{O}C$at room temperature. A galvanometer of 40 ohm 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. ${12}^{O}C$
• B. $0^{O}C$
• C. ${20}^{O}C$
• D. ${16}^{O}C$

Answer 1

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

Temperature difference, $\theta =\frac{V_g}{V}=\frac{R_g{I}_g}{V}=\frac{40\times {10}^{-5}}{25\times {10}^{-6}}=16C$