A platinum resistance temperature sensor has a resistance of 120 Ω at 0℃ and forms one arm of a Wheatstone bridge. At this temperature the bridge is balanced with each of the other arms being 120 Ω. The temperature coefficient of resistance of the platinum is 0.0039/K. What will be the output voltage from the bridge for a change in temperature of 20℃? The loading across the output is effectively open circuit and the supply voltage to the bridge is from a source of 6.0 V with negligible internal resistance.

Question

Jaydin Roman

Engineering

25 November, 18:37

A platinum resistance temperature sensor has a resistance of 120 Ω at 0℃ and forms one arm of a Wheatstone bridge. At this temperature the bridge is balanced with each of the other arms being 120 Ω. The temperature coefficient of resistance of the platinum is 0.0039/K. What will be the output voltage from the bridge for a change in temperature of 20℃? The loading across the output is effectively open circuit and the supply voltage to the bridge is from a source of 6.0 V with negligible internal resistance.

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Answers (1)

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Emmett Warren · Answer 1

Answer : 9.36ohms / temperature

Explanation:

Expression for the variation of resistance of platinum with temperature

Rt = Ro (1+*t)

Rt = resistance @ t°C

Ro = resistance @ 0°C

* = temperature coefficient of resistance

Calculate the change in resistance by putting 120ohms for Ro,

0.0039/K for *

20°C for t

Using this formula:

Rt = Ro (1+*t)

Rt - Ro = Ro*t

= (120ohms) (0.0039/K) (20°C)

= 9.36ohms/K