If `theta_i`, is the inversion temperature, `theta_n` is the neutral temperature, `theta_c` is the temperature of the cold junction, thenA. `theta_(i)+theta_(c)=theta_(n)`B. `theta_(i)-theta_(c)=2 theta_(n)`C. `(theta_(1)+theta_(2))/2 =theta_(n)`D. `theta_(c)-theta_(1)=2 theta_(n)`

If `theta_i`, is the inversion temperature, `theta_n` is the neutral t

1.	If `theta_i`, is the inversion temperature, `theta_n` is the neutral temperature, `theta_c` is the temperature of the cold junction, thenA. `theta_(i)+theta_(c)=theta_(n)`B. `theta_(i)-theta_(c)=2 theta_(n)`C. `(theta_(1)+theta_(2))/2 =theta_(n)`D. `theta_(c)-theta_(1)=2 theta_(n)`
Answer» Correct Answer - C In Seeback (thermoelectric) effect, the temperature of hot junction at which the thermo emf is maximum is called the neutral temperature `(theta_(n))`, and the temperature at which thermo emf changes its sign is called inversion temperature `(theta_(i))`. if `theta_(c)` is the temperature of cold junction then these three are related by the expression. `theta_(n)-theta_(c)=theta_(i)-theta_(n)` `:, theta_(n)=(theta_(i)+theta_(c))/2`

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If `theta_i`, is the inversion temperature, `theta_n` is the neutral temperature, `theta_c` is the temperature of the cold junction, thenA. `theta_(i)+theta_(c)=theta_(n)`B. `theta_(i)-theta_(c)=2 theta_(n)`C. `(theta_(1)+theta_(2))/2 =theta_(n)`D. `theta_(c)-theta_(1)=2 theta_(n)`

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