Question

The thermo emf in lead-iron thermocouple with one junction at $0^{\circ }C$, is given by $e=1784t-{bt}^2$ in volts, where $t^{\circ }C$ is the temperature of the other junction. The neutral temperature is ${371.7}^{\circ }C$. Then the value of $b$ in ${V/\left({}^{\circ }C\right)}^2$ is

• A. $-2.4$
• B. $4.799$
• C. $-9.6$
• D. $+9.6$

Answer 1

Answer: (A)

$-2.4$

$e=0$

$⟹t=0$ or $\frac{1784}{b}$

Hence, inversion temperature $T_i=\frac{1784}{b}$

Neutral temperature, $T_n=\frac{T_i+T_c}{2}$

or, $371.7=\frac{0+1784/b}{2}$

or, $|b|=2.399$

Thermoemf is given by, $e=\alpha t+\beta t^2,\alpha >0,\beta >0$

Hence, comparing with the standard form, we get: $b\approx -2.4$