Current Electricity Charging Discharging Of Capacitors Question 122
Question: The temperature of hot junction of a thermo-couple changes from $ 80^{o}C $ to $ 100^{o}C. $ The percentage change in thermoelectric power is
Options:
A) 8%
B) 10%
C) 20%
D) 25%
Show Answer
Answer:
Correct Answer: D
Solution:
Thermoelectric power $ P\propto \theta $
Therefore $ \frac{P _{100}-P _{80}}{P _{80}}\times 100=\frac{100-80}{80}\times 100 $ = 25%
BETA
