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The EMF generated across a thermocouple ...

The EMF generated across a thermocouple with cold junction at `0^(@)C` , is
`E=a theta+btheta^(2)`
with a `=30muV(.^(@)C)^(-1)` and `b=0.08muV(.^(@)C)^(-2)`
At E=6mV. The hot junction temperature is about

A

`14^(@)C`

B

`144^(@)C`

C

`323^(@)C`

D

`640^(@)C`

Text Solution

Verified by Experts

`E=atheta+btheta^(2)`
It is a quadratic equation
`btheta^(2)+a theta-E=0`
or `theta=(-a+-sqrt(a^(2)+4bE))/(2b)`
Now consider. `sqrt(a^(2)+4bE)` for solution
`sqrt(a^(2)+4bE)`
`=sqrt((30xx10^(-6))^(2)+4 xx0.08xx10^(-6)xx6xx10^(-3))`
`=5.3xx10^(-5)`
`therefore theta=(-3xx10^(-5)+5.3xx10^(-5))/(2xx0.08xx10^(-6))=144^ (@)C`
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