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.
