Question

A copper constantan thermocouple produces an e.m.f. of $40\mu V$ per C . The smallest temperature difference that can be measured with this thermocouple is 2.5 C , when a galvanometer capable of detecting as low as ${10}^{-6}$ amp. is employed. The resistance of that galvanometer is

• A. $50\Omega$
• B. $1000\Omega$
• C. $200\Omega$
• D. $100\Omega$

Answer 1

The correct option is D $100\Omega$

The emf here is given as $e=at$ where $a=40\mu V/{}^{o}C$

So as smallest temperature difference that can be measured is ${2.5}^{o}C$, the smallest emf difference that can be measured is $a\times 2.5=40\times {10}^{-6}\times 2.5={10}^{-4}V$

As the minimum temperature detectable is ${10}^{-6}A$, the galvanometer resistance is ${10}^{-4}/{10}^{-6}=100\Omega$