A thermocouple is immersed in water at 373 K and the other in ice at 273K. The emf of the thermocouple Is 90 (μ)V for each 1 K difference in temperature between junctions, and the thermocouple resistance is 6 Ω. What current will flow in the galvanometer connected in series with an internal resistance of 30 ohms?
Correct answer: B. 250 μA
- A. 1.8 μA
- B. 250 μA
- C. 300 μA
- D. 1.5 mA
- E. 1.8 mA
Explanation
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
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Steady current relates charge flow to potential difference, resistance and resistivity, including how temperature changes a conductor's resistance. Sources with internal resistance have terminal voltage below their emf when delivering current, and maximum power output occurs under a specific load condition rather than at zero resistance.
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