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 :
$\frac{1}{2} k T_{0}$
$\frac{1}{2} k T_{0}^{2}$
$k T_{0}$
$\frac{1}{2} k {(T_{0} - T_{r})}^{2}$

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 :
$\frac{1}{2} k T_{0}$
$\frac{1}{2} k T_{0}^{2}$
$k T_{0}$
$\frac{1}{2} k {(T_{0} - T_{r})}^{2}$

Toppr Admin · Accepted Answer

We know that thermoelectric powerS-dEdTGiven- E-K-T-x2212-Tr-T0-x2212-12-T-Tr-By differentiating the above equation w-r-t- T and putting T-12T0- we get S-12KT0

One junction of a certain thermoelectric couple is at a fixed temperature Tr and the other junction is at temperature T. The thermo-electromotive force for this is expressed by E=k(T−Tr)[T0−12(T+Tr)]. At temperature T=12T0, the thermoelectric power is :12kT012kT20kT012k(T0−Tr)2

We know that thermoelectric powerS=dEdTGiven, E=K(T−Tr)[T0−12(T+Tr)]By differentiating the above equation w.r.t. T and putting T=12T0, we get S=12KT0

We know that thermoelectric power
$S = \frac{d E}{d T}$
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 $T = \frac{1}{2} T_{0}$ , we get $S = \frac{1}{2} K T_{0}$