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QPages 218-219: Q15: HC Verma: Concepts of Physics - Volume 2 Thermal and Chemical Effects of Electric Current Pages 218-219: Q15 Solution: Find the neutral temperature and inversion temperature of a…

⚠️ Common Exam Traps & Conceptual Pitfalls

When calculating the thermocouple coefficients $a$ and $b$ from individual metal-lead data, students frequently fail to subtract the values correctly, leading to sign errors. Remember that $a_{AB} = a_{A} - a_{B}$ and $b_{AB} = b_{A} - b_{B}$; for example, $b_{Cu,Fe} = 0.012 - (-0.030) = 0.042$, not $0.012 + 0.030$ by coincidence of signs, but strictly by subtraction.

Find the neutral temperature and inversion temperature of a copper-iron thermocouple if the reference junction is kept at 0�C. Use the data: For Copper with lead,

a = 2.76 \ \mu\text{V}/^\circ\text{C}

,

b = 0.012 \ \mu\text{V}/^\circ\text{C})^2

. For Iron with lead,

a = 16.6 \ \mu\text{V}/^\circ\text{C}

,

b = -0.030 \ \mu\text{V}/^\circ\text{C})^2

.

Step-by-Step Verified Solution

**Answer:** Neutral temperature = 328.57�C. **Step-by-Step Explanation:**

a_{Cu,Fe} = 2.76 - 16.6 = -13.8\ \mu\text{V}/�\text{C}

.

b_{Cu,Fe} = 0.012 + 0.030 = 0.042\ \mu\text{V}/�\text{C}^2

.

\theta_n = -a/b = 13.8/0.042 = 328.57�\text{C}

.

All Questions in this Chapter

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