Current Electricity Charging Discharging Of Capacitors Question 106
Question: The temperature of the cold junction of thermo‑couple is $ 0{}^\circ C $ and the temperature of hot junction is $ T{}^\circ C $ . The e.m.f. is $ E=16T-0.04T^{2}\mu $ volts. The temperature of inversion is
[EAMCET 1994]
Options:
A) $ 200{}^\circ C $
B) $ 400{}^\circ C $
C) $ 100{}^\circ C $
D) $ 300{}^\circ C $
Show Answer
Answer:
Correct Answer: B
Solution:
Comparing the given equation with $ E=\alpha ,t+\frac{1}{2}\beta ,t^{2} $ We get $ \alpha =16 $ and $ \frac{1}{2}\beta =-0.04,\Rightarrow \beta =-0.08 $
Therefore $ t _{n}=-\frac{\alpha }{\beta }=-\frac{16}{-0.08}=200{}^\circ C $ Also $ t _{i}=2t _{n}-t _{c} $
Therefore $ t _{i}=2\times (200)-0=400{}^\circ C $
BETA
