Question

The thermo-emf of a thermocouple is $25\, \mu V /{ }^{\circ} C$ at room temperature. A galvanometer of $40\, \Omega$ resistance, capable of detecting current as low as $10^{-1} A$, is connected with the thermocoupie. The smallest temperature difference that can be detected by this system is

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

Answer 1

Answer: (A)

Thermo-emf of thermocouple = $-25 \mu V/{}^{\circ} C $
Let $\theta$ be the smallest temperature difference. Therefore, after connecting the thermocouple with the galvanometer, thermo-emf
$E=(25 \mu V/{}^{\circ} C) \times \theta ({}^{\circ} C) $
$=25 \theta \times 10^{-6}V$
Potential drop developed across the galvanometer
$ =iR=10^{-5} \times 40 $
$ =4 \times 10^{-4} V$
$ 4 \times 10^{-4}=25 \theta \times 10^{-6} $
$\therefore \theta= \frac {4}{25} \times 10^2=16^{\circ} C $