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)

Question

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)

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`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`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`