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Chapter 15 · 3 hours

Thermoelectric Effects

NEB Class 12 past exam questions

Past questions and answers

9 questions set from this chapter, 1 of them more than once. Most repeated first.

  • Asked 2 times
  • 2080 Baishakh · 1 mark
  • 2083 Baishakh · 1 mark

Which of the following represents the process of production of emf by maintaining a difference of temperature between the two junctions of a loop of two dissimilar metals?
(a) Seebeck effect (b) Joule effect (c) Peltier effect (d) Thomson effect

Answer

(a) Seebeck effect

When the two junctions of a thermocouple (a loop of two dissimilar metals) are kept at different temperatures, an emf (thermo-emf) is produced and a current flows in the loop. This is the Seebeck effect. The Peltier and Thomson effects are about heat being absorbed or released when a current flows, and Joule heating is I2RI^2R heating.

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  • 2081 Baishakh · 1 mark

The temperature of inversion of a thermocouple is 600∘^\circC and the neutral temperature is 310∘^\circC. What must be the temperature of cold junction?
(A) 0∘^\circC (B) 10∘^\circC (C) 20∘^\circC (D) 30∘^\circC

Similar questions: Cold junction from inversion and neutral temperatures (2082 model)

Answer

(C) 20∘^\circC

The neutral temperature lies midway between the cold-junction and inversion temperatures, θn=θc+θi2\theta_n = \dfrac{\theta_c + \theta_i}{2}, so

θc=2θn−θi=2(310)−600=20∘C\theta_c = 2\theta_n - \theta_i = 2(310) - 600 = 20^\circ\text{C}

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  • 2082 model · 1 mark

For a thermocouple, the inversion temperature is 600∘600^\circC and the neutral temperature is 320∘320^\circC. Find the temperature of the cold junction.
(a) 40∘40^\circC (b) 20∘20^\circC (c) 80∘80^\circC (d) 60∘60^\circC

Similar questions: Cold Junction Temperature (2081 Baishakh)

Answer

(a) 40∘40^\circC

The neutral temperature is the mean of the cold-junction and inversion temperatures:

θn=θc+θi2  ⇒  320=θc+6002  ⇒  θc=640−600=40∘C\theta_n = \frac{\theta_c + \theta_i}{2} \;\Rightarrow\; 320 = \frac{\theta_c + 600}{2} \;\Rightarrow\; \theta_c = 640 - 600 = 40^\circ\text{C}

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 Jestha · 2+3 marks

(a) Differentiate Seebeck's effect and Peltier's effect. [2]
(b) Explain the variation of thermo-emf with temperature. [3]

Answer

(a) Seebeck effect vs Peltier effect

Seebeck effectPeltier effect
When the two junctions of a thermocouple are kept at different temperatures, an emf is produced in the circuit.When a current is passed through a junction of two dissimilar metals, heat is absorbed or released at the junction.
Heat energy is converted into electrical energy.Electrical energy causes heating or cooling at a junction (heat is pumped).
The temperature difference is the cause and the emf is the effect.The current is the cause and the heating/cooling is the effect.
Used in thermocouples, thermometers and thermoelectric generators.Used in thermoelectric refrigerators and coolers.

The Peltier effect is reversible: reversing the current changes heating to cooling at the junction.

(b) Variation of thermo-emf with temperature

Keep the cold junction at 0∘0^\circC and slowly raise the temperature θ\theta of the hot junction. Experiments show that the thermo-emf varies as

E=aθ+12bθ2E = a\theta + \frac{1}{2}b\theta^2

where aa and bb are constants for the pair of metals (bb is negative). The graph of EE against θ\theta is a parabola.

   E
   ^
   |              .  -  -  .
   |          .'      :      '.
   |        /         :         \
   |      /           :           \
   |    /             :             \
   |  /               :               \
   +------------------+-----------------+-----> θ
  θc (0 °C)           θn                θi
                   (neutral)       (inversion)
  1. Rise: As θ\theta increases, EE increases, but less and less quickly.
  2. Neutral temperature θn\theta_n: EE reaches its maximum. Here
dEdθ=a+bθn=0⇒θn=−ab\frac{dE}{d\theta} = a + b\theta_n = 0 \quad\Rightarrow\quad \theta_n = -\frac{a}{b}

θn\theta_n depends only on the pair of metals, not on the temperature of the cold junction. 3. Fall and inversion temperature θi\theta_i: Beyond θn\theta_n the emf decreases and becomes zero at θi\theta_i. Above θi\theta_i the emf reverses its direction. 4. Because the parabola is symmetric about θn\theta_n,

θn−θc=θi−θn⇒θn=θi+θc2\theta_n - \theta_c = \theta_i - \theta_n \quad\Rightarrow\quad \boxed{\theta_n = \frac{\theta_i + \theta_c}{2}}

For example, for a Cu–Fe thermocouple with the cold junction at 0∘0^\circC, θn≈270∘\theta_n \approx 270^\circC and θi≈540∘\theta_i \approx 540^\circC.

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  • 2079 Jestha · 1 mark

If the emf of a thermocouple, one junction of which is kept at 0∘^\circC is given by E=aθ+12bθ2E = a\theta + \dfrac{1}{2}b\theta^2. Then, the neutral temperature will be
(a) ab\dfrac{a}{b} (b) −ab-\dfrac{a}{b} (c) 1ab\dfrac{1}{ab} (d) −1ab-\dfrac{1}{ab}

Answer

(b) −ab-\dfrac{a}{b}

At the neutral temperature the emf is maximum, so dEdθ=0\dfrac{dE}{d\theta} = 0:

a+bθn=0⇒θn=−aba + b\theta_n = 0 \quad\Rightarrow\quad \theta_n = -\frac{a}{b}

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  • 2082 Baishakh · 3 marks

a) Explain the variation of thermoelectric emf with the temperature of the hot junction of a thermocouple. [3]

Answer

Keep the cold junction of a thermocouple (e.g. Cu–Fe) at a fixed temperature θc\theta_c (say 0∘0^\circC) and gradually heat the hot junction to temperature θ\theta. The thermo-emf is found to vary as

E=aθ+12bθ2E = a\theta + \frac{1}{2}b\theta^2

where aa and bb are thermoelectric constants of the pair (bb is negative). The EE–θ\theta graph is a parabola.

   E
   ^
   |              .  -  -  .
   |          .'      :      '.
   |        /         :         \
   |      /           :           \
   |    /             :             \
   |  /               :               \
   +------------------+-----------------+-----> θ
  θc                  θn                θi
                   (neutral)       (inversion)
  1. Increase: At first EE increases with θ\theta, but its rate of increase gets smaller.
  2. Neutral temperature (θn\theta_n): This is the hot-junction temperature at which the thermo-emf is maximum. Setting dEdθ=a+bθ=0\dfrac{dE}{d\theta} = a + b\theta = 0 gives θn=−ab\theta_n = -\dfrac{a}{b}. It is a constant for a given pair of metals and does not depend on θc\theta_c.
  3. Decrease and inversion temperature (θi\theta_i): Above θn\theta_n the emf decreases and becomes zero at the inversion temperature θi\theta_i. Beyond θi\theta_i the emf changes sign (the current reverses).
  4. The curve is symmetric about θn\theta_n:
θn=θc+θi2\theta_n = \frac{\theta_c + \theta_i}{2}

For Cu–Fe with θc=0∘\theta_c = 0^\circC: θn≈270∘\theta_n \approx 270^\circC and θi≈540∘\theta_i \approx 540^\circC.

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  • 2080 Baishakh · 1+2+1+1 marks

The figure shows the variation of thermoemf with the temperature of the hot junction, when the cold junction is at 0°C.
i) What is thermoelectric effect? [1]
ii) Define θn\theta_n and θi\theta_i. [2]
iii) What will be the effect on θi\theta_i if we shift the cold junction to −10∘-10^\circC? [1]
iv) Give one of the applications of such thermocouple. [1]
Figure: emf against temperature of the hot junction: a parabola starting at θc\theta_c on the axis, peaking at θn\theta_n and returning to the axis at θi\theta_i.

Answer

(i) Thermoelectric effect

When the two junctions of a closed circuit of two different metals are kept at different temperatures, an emf (thermo-emf) is produced and a current flows in the circuit. This is the Seebeck effect.

(ii) θn\theta_n and θi\theta_i

  • θn\theta_n is the neutral temperature: the temperature of the hot junction at which the thermo-emf is maximum.
  • θi\theta_i is the temperature of inversion: the temperature of the hot junction at which the thermo-emf becomes zero and then reverses its direction.

(iii) Effect of shifting the cold junction to −10∘-10^\circC

The neutral temperature stays the same, and θi=2θn−θc\theta_i = 2\theta_n - \theta_c, so θi\theta_i increases by 10∘10^\circC.

(iv) Application

It can be used as a thermocouple thermometer to measure temperatures (for example of furnaces), and a combination of thermocouples forms a thermopile to detect radiant heat.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • NEB model question 2079, Physics 1021 · 1 mark

The neutral temperature of a thermocouple is equal to 500∘500^\circC when the temperature of cold junction is 0∘0^\circC. Percentage change in the temperature of inversion when temperature of cold junction is equal to 20∘20^\circC is
(A) 2% (B) 3% (C) 4% (D) 5%

Answer

(A) 2%

Using θi=2θn−θc\theta_i = 2\theta_n - \theta_c, and noting that θn=500∘\theta_n = 500^\circC does not change with θc\theta_c:

  • with θc=0∘\theta_c = 0^\circC: θi=1000∘\theta_i = 1000^\circC
  • with θc=20∘\theta_c = 20^\circC: θi=980∘\theta_i = 980^\circC
Percentage change=1000−9801000×100%=2%\text{Percentage change} = \frac{1000 - 980}{1000} \times 100\% = 2\%

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  • CDC model question 2078, Physics · 1 mark

What happens to the neutral temperature if the cold junction of a thermocouple is decreased?
(a) increases (b) decreases (c) remains the same (d) approaches inversion temperature

Answer

(c) remains the same

The neutral temperature θn=−ab\theta_n = -\dfrac{a}{b} depends only on the two metals of the thermocouple, not on the cold-junction temperature. Changing the cold junction changes only the inversion temperature, since θi=2θn−θc\theta_i = 2\theta_n - \theta_c.

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Questions from NEB Class 12 2079 Physics, 1021 'O' (Board exam, Jestha 2079, the set used outside Bagmati (typed copy)), NEB Class 12 2079 Physics, 1021 'V' (Board exam, Jestha 2079, the Bagmati set (typed copy)), NEB Class 12 2080 Physics, 1021 'Q' (Board exam, Baishakh 2080 (scan)), NEB Class 12 2081 Physics, 1021 'C' (Board exam, Baishakh 2081 (scan)), NEB Class 12 2082 Physics, 1021 'H' (Board exam, Baishakh 2082 (scan)), NEB Class 12 2083 Physics, 1021 'E' (Board exam, Baishakh 2083 (scan)), NEB model question 2079, Physics 1021 (NEB's model paper for the 2080 exam), CDC model question 2078, Physics (CDC's model paper, in the Grade 12 specification grid), NEB Class 12 2080 Physics 1021 (nebexam.com 'Old Question 2080', same as its 'Model Question Set B') (Board-style paper typed on nebexam.com as the 2080 old question, 75 marks; not the Q-set already on the site (neb-2080-q)), NEB Class 12 Physics 1021 Model Question 2082 (NEB model paper, 2082 (typed copy on nebexam.com)) and NEB Class 12 Physics Model Question 2079, Set I (NEB model paper, Set I, 2079 (questions typed at the foot of a solutions post; solutions not used)). Answers are written for this site; check them against your class notes.

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