The thermo e.m.f. of a thermocouple is 25,mu | vedclass

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(B) The thermo $e.m.f.$ $(e)$ produced by the thermocouple is given by the product of the thermo $e.m.f.$ coefficient $(\alpha)$ and the temperature d

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) The thermo $e.m.f.$ $(e)$ produced by the thermocouple is given by the product of the thermo $e.m.f.$ coefficient $(\alpha)$ and the temperature difference $(\Delta 	heta)$: $e = \alpha \Delta 	heta$.According to Ohm's law, the $e.m.f.$ is also equal to the product of the current $(i)$ and the resistance $(R)$: $e = iR$.Equating the two expressions: $\alpha \Delta 	heta = iR$.Given: $\alpha = 25\,\mu V/^{\circ}C = 25 	imes 10^{-6}\,V/^{\circ}C$, $i = 10^{-5}\,A$, and $R = 40\,\Omega$.Substituting the values: $(25 	imes 10^{-6}) 	imes \Delta 	heta = 10^{-5} 	imes 40$.$\Delta 	heta = \frac{40 	imes 10^{-5}}{25 	imes 10^{-6}} = \frac{400}{25} = 16\,^{\circ}C$.Therefore, the smallest temperature difference that can be detected is $16\,^{\circ}C$.

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