GATE 2022 IN Q.19: solution and answer

GATE 2022 · INQ.19 · 1 mark · MCQ

The figure shows a Chromel-Alumel thermocouple, where the junction A is held at temperature $T_{A}$ , and a thermal emf $E_{1}$ is measured using an ideal voltmeter between the open ends B1 and B2, both held at temperature $T_{B}$ . Two identical copper wires are introduced between B1-C1 and B2-C2 as shown in the figure. When C1 and C2 are held at temperature $T_{C}$ , the voltmeter reads a thermal emf $E_{2}$ . Then, ________

GATE 2022 IN question 19: Two sketches. — Figure from the GATE 2022 IN paper. It is part of the question.

(A)
$E_{1} < E_{2}$
(B)
$E_{1} > E_{2}$
(C)
$E_{1} = 2 E_{2}$
(D)
$E_{1} = E_{2}$

Answer and solution

ANSWER

$E_{1} = E_{2}$ , because the copper extension wires and the equal-temperature junctions add no net emf. This is option (D).

Official answer: DSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Top circuit: the voltmeter is connected across B1 and B2, both at $T_{B}$ , so it reads the Chromel-Alumel emf between the hot junction A at $T_{A}$ and the open ends at $T_{B}$ : $E_{1} = e_{C A} (T_{A}, T_{B})$ .
Bottom circuit: the loop is Chromel and Alumel meeting at A ( $T_{A}$ ), then a Chromel-copper junction at B1 and an Alumel-copper junction at B2, both at $T_{B}$ . Two identical copper wires then run from B1 to C1 and from B2 to C2, and the voltmeter is connected across C1 and C2, both at $T_{C}$ .
Law of homogeneous material: a wire of one uniform metal produces no net emf from a temperature gradient along it. Each copper wire runs from $T_{B}$ to $T_{C}$ , so neither adds an emf. Because the two copper wires are identical, any small thermoelectric property they have is the same, and it cancels around the loop.
Law of intermediate metals: inserting a third metal (copper) into the circuit changes nothing if its two junctions with the thermocouple metals are at the same temperature. Both copper junctions, Chromel-Cu at B1 and Alumel-Cu at B2, are at $T_{B}$ , so their emfs are equal and opposite and cancel: $e_{Cr-Cu} (T_{B}) - e_{Al-Cu} (T_{B})$ is exactly the emf that a direct Chromel-Alumel contact at $T_{B}$ would give, so it is already part of the reference end of $E_{1}$ .
The voltmeter terminals at C1 and C2 are both at $T_{C}$ . The two copper-to-instrument junctions are at the same temperature, so they also cancel, and the reading does not depend on $T_{C}$ .
Hence the voltmeter reads $E_{2} = e_{C A} (T_{A}, T_{B}) = E_{1}$ , whatever the value of $T_{C}$ . Option (D) is correct.

KEY IDEA

Extra wires of a third metal do not change a thermocouple's emf if the two junctions with it are at the same temperature (law of intermediate metals). A homogeneous wire carrying a temperature gradient also adds no emf (law of homogeneous material). The reading depends only on the measuring junction temperature and the reference junction temperature $T_{B}$ .

WATCH OUT FOR

Thinking the temperature $T_{C}$ at the voltmeter end becomes a new reference junction, so that the reading changes to depend on $T_{C}$ (for example $E_{2} = e_{C A} (T_{A}, T_{C})$ ). In fact the reference stays at $T_{B}$ , because copper is a third metal whose two junctions are at the same temperature.

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 5 Temperature questions, with notes

GATE 2022 IN Q.19: solution and answer

Answer and solution

More Temperature questions