A thermocouple of negligible resistance produ | vedclass

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(C) The sensitivity of the galvanometer is $1\,\mu A/	ext{div}$. To detect the smallest temperature difference,we need a minimum deflection of $1\,

Vedclass · Accepted Answer

$0.25$

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(C) The sensitivity of the galvanometer is $1\,\mu A/	ext{div}$. To detect the smallest temperature difference,we need a minimum deflection of $1\,	ext{div}$.Therefore,the minimum current required is $I = 1\,\mu A$.The resistance of the galvanometer is $R = 10\,\Omega$.The voltage (e.m.f.) required to produce this current is $V = I 	imes R = 1\,\mu A 	imes 10\,\Omega = 10\,\mu V$.Given that the thermocouple produces an $e.m.f.$ of $40\,\mu V/^{\circ}C$,the temperature difference $\Delta T$ corresponding to $10\,\mu V$ is calculated as:$\Delta T = \frac{V}{	ext{sensitivity of thermocouple}} = \frac{10\,\mu V}{40\,\mu V/^{\circ}C} = 0.25^{\circ}C$.Thus,the smallest temperature difference that can be detected is $0.25^{\circ}C$.

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