The emf of a thermocouple, with one junction | vedclass

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(A) The Peltier coefficient $\pi$ is defined as $\pi = T \frac{de}{dT}$, where $T$ is the absolute temperature in Kelvin.Given the temperature in Cels

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The Peltier coefficient $\pi$ is defined as $\pi = T \frac{de}{dT}$, where $T$ is the absolute temperature in Kelvin.Given the temperature in Celsius is $t$, the absolute temperature is $T = t + 273$.Substituting $t = T - 273$ into the given equation $e = at + bt^2$:$e = a(T - 273) + b(T - 273)^2$Now, differentiate $e$ with respect to $T$:$\frac{de}{dT} = a + 2b(T - 273)$Since $T - 273 = t$, we have $\frac{de}{dT} = a + 2bt$.Substituting this into the formula for the Peltier coefficient:$\pi = T \frac{de}{dT} = (t + 273)(a + 2bt)$.

List-$I$	List-$II$
$I$. Joule	$A$. $\text{Henry} \times \text{Amp/sec}$
$II$. Watt	$B$. $\text{Farad} \times \text{Volt}$
$III$. Volt	$C$. $\text{Coulomb} \times \text{Volt}$
$IV$. Coulomb	$D$. $\text{Oersted} \times \text{cm}$
	$E$. $\text{Amp} \times \text{Gauss}$
	$F$. $\text{Amp}^2 \times \text{Ohm}$

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

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