Question

The thermo e.m. $f$. of a thermo-couple 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. ${16}^{o}C$
• B. ${12}^{o}C$
• C. $8^{o}C$
• D. ${20}^{o}C$

Answer 1

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

Given : $R=40$ ohm $I={10}^{-5}$ A

Voltage across the galvanometer $V=IR={10}^{-5}\times 40=4\times {10}^{-4}$ volts

$\therefore$ Smallest temperature difference to be detected $\theta =\frac{4\times {10}^{-4}}{25\times {10}^{-6}}={16}^o$ C