The emf of a thermocouple,one junction of whi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The Peltier coefficient $\pi$ is given by the relation $\pi = T \frac{de}{dT}$,where $T$ is the absolute temperature in Kelvin.Given that one junc

Vedclass · Accepted Answer

$(t + 273)(a + 2bt)$

Vedclass · Answer

(A) The Peltier coefficient $\pi$ is given by the relation $\pi = T \frac{de}{dT}$,where $T$ is the absolute temperature in Kelvin.Given that one junction is at $0\,^{\circ}C$,the absolute temperature $T$ is related to the Celsius temperature $t$ by $T = t + 273$.Substituting $t = T - 273$ into the given emf equation $e = at + bt^2$:$e = a(T - 273) + b(T - 273)^2$.Differentiating with respect to $T$:$\frac{de}{dT} = a + 2b(T - 273)$.Since $T - 273 = t$,we have $\frac{de}{dT} = a + 2bt$.Now,calculating the Peltier coefficient:$\pi = T \frac{de}{dT} = (t + 273)(a + 2bt)$.

The emf of a thermocouple,one junction of which is kept at $0\,^{\circ}C$,is given by $e = at + bt^2$. The Peltier coefficient will be

Similar Questions

The electrochemical equivalents of $Cu$ and $Ag$ are $7 \times 10^{-6}$ and $1.2 \times 10^{-6}$ respectively. If a certain current deposits $14 \ g$ of $Cu$,then the amount of $Ag$ deposited is .......... $g$.

The unit of $e.m.f.$ is

In the given circuit,the value of $\left|\frac{I_1+I_3}{I_2}\right|$ is

In the circuit shown,the galvanometer $G$ of resistance $60 \, \Omega$ is shunted by a resistance $r = 0.02 \, \Omega$. The current through $R$ is nearly $1 \, A$. The value of resistance $R$ (in $\Omega$) is nearly:

Find the current in the circuit. (in $ A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module