For a certain thermocouple the emf is $E = aT + bT^{2}$, where $t\ (in℃)$ is the temperature of hot junction, the cold junction is at 0$℃$. The value of contants $a\ and\ b$ are $10 imes 10^{- 6}$ and $0.02 imes 10^{- 6}$ respectively, then the temperature of inversion $(in℃)$ will be

Question

For a certain thermocouple the emf is $E = aT + bT^{2}$, where $t\ (in℃)$ is the temperature of hot junction, the cold junction is at 0$℃$. The value of contants $a\ and\ b$ are $10 	imes 10^{- 6}$ and $0.02 	imes 10^{- 6}$ respectively, then the temperature of inversion $(in℃)$ will be

Dr. Abubacker · Accepted Answer

(c) $E = aT + bT^{2}$
At temperature of inversion, $E = 0,$
$	herefore aT_{i} + bT_{i}^{2} = 0$
$\Rightarrow T_{i} = - \frac{a}{b}$
$\Rightarrow T_{i} = - \frac{10 	imes 10^{- 6}}{(0.02 	imes 10^{- 6})} = 500℃$

Q. For a certain thermocouple the emf is $E = aT + bT^{2}$, where $t\ (in℃)$ is the temperature of hot junction, the cold junction is at 0$℃$. The value of contants $a\ and\ b$ are $10 \times 10^{- 6}$ and $0.02 \times 10^{- 6}$ respectively, then the temperature of inversion $(in℃)$ will be

Explanation

Reviewed by Dr. Abubacker

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