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