Current Electricity Charging Discharging Of Capacitors Question 119
Question: The emf of a thermocouple, one junction of which is kept at $ 0^{o}C, $ is given by $ e=at+bt^{2} $ the Peltier co-efficient will be
Options:
A) $ (t+273)(a+2bt) $
B) $ (t+273)(a-2bt) $
C) $ (t-273)(a-2bt) $
D) $ (t-273)(a-2bt) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \because $ Peltier coefficient $ \pi =T\frac{de}{dT} $ and $ t^{o}C=T-273 $ \ $ e=a(T-273)+b{{(T-273)}^{2}} $ Differentiating w.r.t. T $ \frac{de}{dT}=a+2b(T-273) $
$ \pi =T\frac{de}{dT}=T[a+2b(T-273)] $
Therefore $ \pi =(t+273),(a+2bt) $
BETA
