Current Electricity Charging Discharging Of Capacitors Question 103
Question: The expression for thermo e.m.f. in a thermocouple is given by the relation $ E=40,\theta -\frac{{{\theta }^{2}}}{20} $ , where $ \theta $ is the temperature difference of two junctions. For this, the neutral temperature will be
[AMU (Engg.) 2000]
Options:
A) $ 100{}^\circ C $
B) $ 200{}^\circ C $
C) $ 300{}^\circ C $
D) $ 400{}^\circ C $
Show Answer
Answer:
Correct Answer: D
Solution:
Comparing the given equation with standard equation $ E=\alpha t+\frac{1}{2}\beta t^{2} $
$ \alpha =40 $ and $ \frac{1}{2}\beta =-\frac{1}{20} $
Therefore $ \beta =-\frac{1}{10} $ Hence neutral temperature $ t _{n}=-\frac{\alpha }{\beta }=\frac{-,40}{-1/10} $
Therefore $ t _{n}=400^{o}C $
BETA
