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If the total emf in a thermocouple is a parabolic function expressed as $\mathrm{E}=$ at $+\frac{1}{2} \mathrm{bt}^{2}$, which of the following relations does not hold good
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The correct answer is:
$t_{n}=a / b$
$\mathrm{E}=\mathrm{at}+\frac{1}{2} \mathrm{bt}^{2}$ is the parabolic equation for thermo emf.
The thermoelectric power is
$\mathrm{S}=\frac{\mathrm{dE}}{\mathrm{dt}}$
$\Rightarrow \mathrm{S}=\mathrm{a}+\mathrm{bt}$. The graph between $\mathrm{S}\left(=\frac{\mathrm{dE}}{\mathrm{dt}}\right)$ and straight line.
When $\mathrm{t}=0, \mathrm{~S}=\mathrm{a}$ (intercept).
At neutral temperature $\frac{\mathrm{dE}}{\mathrm{dt}}=0$ and
$\mathrm{t}=\mathrm{t}_{\mathrm{n}} \Rightarrow 0=\mathrm{a}+\mathrm{bt}_{\mathrm{n}} \Rightarrow \mathrm{t}_{\mathrm{n}}=\frac{-\mathrm{a}}{\mathrm{b}}$ and at cold junction $\mathrm{t}_{\mathrm{i}}=2 \mathrm{t}_{\mathrm{n}}=\frac{-2 \mathrm{a}}{\mathrm{b}}$
The thermoelectric power is
$\mathrm{S}=\frac{\mathrm{dE}}{\mathrm{dt}}$
$\Rightarrow \mathrm{S}=\mathrm{a}+\mathrm{bt}$. The graph between $\mathrm{S}\left(=\frac{\mathrm{dE}}{\mathrm{dt}}\right)$ and straight line.
When $\mathrm{t}=0, \mathrm{~S}=\mathrm{a}$ (intercept).
At neutral temperature $\frac{\mathrm{dE}}{\mathrm{dt}}=0$ and
$\mathrm{t}=\mathrm{t}_{\mathrm{n}} \Rightarrow 0=\mathrm{a}+\mathrm{bt}_{\mathrm{n}} \Rightarrow \mathrm{t}_{\mathrm{n}}=\frac{-\mathrm{a}}{\mathrm{b}}$ and at cold junction $\mathrm{t}_{\mathrm{i}}=2 \mathrm{t}_{\mathrm{n}}=\frac{-2 \mathrm{a}}{\mathrm{b}}$
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