Question

Antimony - Bismuth thermocouple generates a thermo emf of 100μV/0C When the cold junction is at 00C . A micro ammeter of resistance 100 Ω is connected in the circuit of Sb - Bi thermocouple. If the least count of the micro ammeter is 1μA. The minimum temperature difference it can measure is (Neglect the resistance of Sb - Bi thermocouple.)
(a)${{1}^{\circ }}C$
(b)${{2}^{\circ }}C$
(c)${{3}^{\circ }}C$
(d)${{4}^{\circ }}C$

Answer 1

In this question we have to first calculate the voltage using the given emf value. Now after calculating the voltage we can use the temperature voltage density to calculate the temperature.

Step by step solution,
The minimum voltage that should be generated by thermocouple so that ammeter senses the change in current is
$V=0.1\mu A\times 100\Omega =100\mu V$
so minimum temperature difference across thermocouple should be
$\Delta T=100\mu V/100\mu {{V}^{\circ }}C-1={{1}^{\circ }}C$

Thus, the correct answer to this question is option (a).

Additional Information: Bismuth comes before Antimony in the thermoelectric series. At the cold junction, current flows from metal that is later in thermoelectric series to the one that is earlier in the thermoelectric series.The thermoelectric effect is the direct conversion of temperature differences to electric voltage and vice versa via a thermocouple.A thermoelectric device creates a voltage when there is a different temperature on each side. Conversely, when a voltage is applied to it, heat is transferred from one side to the other, creating a temperature difference. At the atomic scale, an applied temperature gradient causes the charge carrier in the material to diffuse from the hot side to the cold side.

Note: While solving this question, one should be familiar with the concept of emf and voltage. We have to calculate the emf and then temperature using the emf temperature relation given in the question.