1. Thermoelectric emf relation

For a thermocouple, the emf between hot junction at temperature $\theta$ and cold junction at temperature $\theta_c$ is written as

$E = a(\theta-\theta_c) + \frac{1}{2}b(\theta^2-\theta_c^2).$

2. Neutral temperature

The neutral temperature $\theta_n$ is the hot junction temperature at which emf is maximum. So,

$\frac{dE}{d\theta} = a + b\theta = 0.$

Hence,

$\theta_n = -\frac{a}{b}.$

3. Inversion temperature

The inversion temperature $\theta_i$ is the hot junction temperature at which emf becomes zero again. So set $E=0$:

$a(\theta_i-\theta_c) + \frac{1}{2}b(\theta_i^2-\theta_c^2)=0.$

Factorizing,

$(\theta_i-\theta_c)\left[a+\frac{1}{2}b(\theta_i+\theta_c)\right]=0.$

Ignoring the trivial root $\theta_i=\theta_c$, we get

$a+\frac{1}{2}b(\theta_i+\theta_c)=0.$

So,

$\theta_i+\theta_c = -\frac{2a}{b}.$

But since

$\theta_n = -\frac{a}{b},$

therefore,

$\theta_i+\theta_c = 2\theta_n.$

Hence,

$\boxed{\frac{\theta_i+\theta_c}{2}=\theta_n}.$

4. Checking options

• A: $\theta_i+\theta_c=\theta_n$ ❌
• B: $\theta_i-\theta_c=2\theta_n$ ❌
• C: $\dfrac{\theta_i+\theta_c}{2}=\theta_n$ ✅
• D: $\theta_c-\theta_i=2\theta_n$ ❌

Therefore, the correct option is C.