If a transmitter outputs 14.4 mA in a 4-20 mA loop that spans 0-100% signal, what percent is the process variable?

• A. 75%
• B. 65%
• C. 50%
• D. 35%

Answer: (B)

In a 4-20 mA loop, 4 mA represents 0% and 20 mA represents 100% of the process variable. To convert the measured current to a percent, use: percent = (I − 4) / (20 − 4) × 100%. For I = 14.4 mA, the difference from the minimum is 14.4 − 4 = 10.4 mA, and the span is 20 − 4 = 16 mA. So 10.4 / 16 = 0.65, which is 65%. Therefore, the process variable is 65%. A quick check: each 0.16 mA corresponds to 1% (since 16 mA equals 100%), and 10.4 mA above the minimum equals 10.4 / 0.16 = 65%.