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Linear Relationships for 4-20 mA transmitters

Linear Relationships for 4-20 Transmitters

A standard electronic pressure transmitter has a range of 50 to 250 psig (345 to 1724 kPa gage), and the pressure is 150 psig (1034 kPa gage), what is the output in mA?

a.   4.0 mA

b.   12.0 mA

c.   13.6 mA

d.   50.0 mA

 

The formula I use for the above question is as follows:

((Input – Input Minimum) / (Input Max –Input Min) * 16)) + 4 = mA

 

Therefore 150-50 =100

And           250-50=200

100 divided by 200 = .5

.5 * 16 = 8

And 8 + 4 = 12 mA

 

If the milliamps were given, this formula would solve for the input.

(((mA Output – 4 mA) / (mA MAX – mA MIN)) * (Input Max – Input Min)) + Input Min = Input

Therefore 12-4 = 8

And 20-4 = 16

8 divide by 16 = .5

250-50 = 200

.5 * 200 = 100

100 + 50 = 150 psig

 

The above formulas are for linear transmitters and URV= Input Max and LRV=Input MIN.

The answer is B.

Mark Weisner
Mark Weisner
Mark Weisner is a technical specialist at ISA.

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