Current Electricity and Electrical InstrumentMedium
Question
A thermocouple of negligible resistance produces an e.m.f of 40 μ V/oC in the linear range of temperature. A galvanometer of resistance 10 Ω whose sensitivity is 1μ A/div, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be
Options
A.0.25oC
B.0.5oC
C.1oC
D.0.1oC
Solution
Thermoemf of thermocouple = 40μ V/oC
Let θoC be smallest temperature difference that can be measured.
∴ After connecting the thermocouple with the galvanometer, thermoemf,
ε = 10 μ V/oC × θoC = 40θo μV
Smallest potential difference developed across the galvanometer
= IR = 1μA × 10 Ω = 10μV
∴ 40θo μV = 10μV
θ =
oC = 0.25o C
Let θoC be smallest temperature difference that can be measured.
∴ After connecting the thermocouple with the galvanometer, thermoemf,
ε = 10 μ V/oC × θoC = 40θo μV
Smallest potential difference developed across the galvanometer
= IR = 1μA × 10 Ω = 10μV
∴ 40θo μV = 10μV
θ =
oC = 0.25o CCreate a free account to view solution
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