Thermoelectric constant of a thermocouple are `alpha` and `beta`. Thermoelectric power at inversion tempreture is

Options

A. `alpha`
B. `- alpha`
C. `(alpha)/(beta)`
D. `-(alpha)/(beta)`

Answer

B

Explanation

`E = alpha T + 1/3 beta T^(2)`
For inversion temperature
`E = 0, T_(i) = -(2alpha)/(beta)`
Thermoelectric power ltbr. `P = (dE)/(dT) = alpha + 1/2 beta.(2T) = alpha+beta T_(i)`
`alpha+beta(-(2alpha)/(beta))=-alpha`.