The thermo-emf of a thermocouple is (25 \mu V/^\circ C) at r... | Physics | SolveeIt | Solveeit

Today, August 23

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

$16 ^\circ C$

Explanation

Solution

Thermo-emf of thermocouple = $-25 \mu V/^\circ C$ Let $?$ be the smallest temperature difference. Therefore, after connecting the thermocouple with the galvanometer, thermo-emf $\hspace5mm E=(25 \mu V/^\circ C) \times ? (^\circ C)$ $\hspace5mm =25 ? \times 10^{-6}V$ Potential drop developed across the galvanometer $\hspace5mm =iR=10^{-5} \times 40$ $\hspace5mm =4 \times 10^{-4} V$ $\therefore \hspace15mm 4 \times 10^{-4}=25 ? \times 10^{-6}$ $\therefore \hspace15mm ?= \frac {4}{25} \times 10^2=16^\circ C$