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NEED HELP>>>>>> - QuestionCove
OpenStudy (prakharluv):

NEED HELP>>>>>>

4 years ago
OpenStudy (prakharluv):

The e.m.f E of certain thermocouple depends on the temperature difference θ between it two junction in accordance with the relation E=70θ-θ^2/20 where E is in micro volt, θ in degree Celsius and one junction is at zero degree Celsius. If E may be determined to ± 100μV, the possible error in ℃ when measuring a temperature of 200℃ is????

4 years ago
OpenStudy (theeric):

I'm not familiar with this stuff, but I'll give you a source and my thoughts. I wouldn't count on me being correct, but maybe it will jog your memory! http://ibguides.com/physics/notes/measurement-and-uncertainties See sections 1.2.10 and 1.2.11. \(E=10\ \theta-\dfrac{\theta^2}{20}\) For example of the errors, Absolute error of \(E\) is \(\pm100\ [\mu V]\). Fractional error (a.k.a. relative error) of \(E\) is \(\dfrac{100\ [\mu V]}{E}\). That error of the other side of the equation will be the sum of the absolute errors of each term. I'll use the \(\Delta\) symbol for uncertainty. \(\Delta E=\Delta(10\ \theta)+\Delta\left(\dfrac{\theta^2}{20}\right)\) The absolute uncertainty of \(\Delta(10\ \theta)\), since \(10\) is exact, is \(\Delta\theta\). If the absolute error of \(\theta\) is \(x_\theta\), then the error of that term is \(x_theta\). The absolute uncertainty of \(\Delta\left(\dfrac{\theta^2}{20}\right)\), since \(\frac{1}{20}\) is exact, is \(\Delta\left(\theta^2\right)\). And \(\Delta\left(\theta^2\right)\) is the two times the fractional error of \(\theta\). So the error of this term, if absolute error is \(x_\theta\), is \(2\left(\dfrac{x_\theta}{\theta}\right)\). So, if we say the absolute error of \(\theta\) is \(x_\theta\) and absolute error of \(E\) is \(x_E\), then\[x_E=x_\theta+2\left(\dfrac{x_\theta}{\theta}\right)\] We know \(x_E=100\ [\mu V]\), and we know \(\theta=(200-0)\ [\ ^\circ C]\). We would need to find \(x_\theta\), if I am correct. Solving for \(x_theta\) will not be too difficult for someone good at algebra. Again, I am unfamiliar with this material, but I hope I have helped. Best of luck! :)

4 years ago
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