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
$=iR=10^{-5}\times 40$
$=4\times 10^{-4}V$
$4\times 10^{-4}=25\theta \times 10^{-6}$
$\therefore \theta =\frac{4}{25}\times 10^2=16^{\circ }C$