Question

A Thermocouple is connected across a galvanometer of resistance 30 ohm. One junction is immersed in water at 373K and the other in ice at 273k. The emf of the thermocouple is 90microvolt for each 1K differece in temperature between the junctions and the thermocouple resistance is 6ohm. What current will flow in the galvanometer?

A) 1.8micro ampere
B) 250micro ampere
C) 300micro ampere
D) 1.5micro ampere
E) 1.8 micro ampere

Answer 1

Given details:

Galvanometer resistance = 30 ohm

Thermocouple couple resistance = 6 ohm

So, Total resistance = 30+6 = 36 ohm

Temperature difference = 373K – 273K =100K

Emf of thermocouple for 1K temperature difference = 90 microvolt = 90 × 10^-6 V

So, for 100K, emf  = 90 × 10^-6 x 100 = 90 × 10^-4 V

Solution:

Current flow in galvanometer = Voltage / Resistance

I = 90 × 10^-4 / 36 = 0.00025 A = 250 micro Ampere