Current Electricity Charging Discharging Of Capacitors Question 105
Question: One junction of a certain thermoelectric couple is at a fixed temperature $ T _{r} $ and the other junction is at temperature T. The thermo electromotive force for this is expressed by $ E=K(T-T _{r})
[ T _{0}-\frac{1}{2}(T+T _{r}) ] $ . At temperature $ T=\frac{1}{2}T _{0} $ , the thermoelectric power is [MP PMT 1994]
Options:
A) $ \frac{1}{2}KT _{0} $
B) $ KT _{0} $
C) $ \frac{1}{2}KT _{0}^{2} $
D) $ \frac{1}{2}K{{(T _{0}-T _{r})}^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
We know that thermoelectric power $ S=\frac{dE}{dT} $ Given $ E=k,(T-T _{r}),[ T _{0}-\frac{1}{2}(T+T _{r}) ] $ By differentiating the above equation w.r.t. T and Putting $ T=\frac{1}{2}T _{o}, $ we get $ S=\frac{1}{2}kT _{o} $
BETA
