May I ask everyone how to solve this problem?
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The voltage of the differential pressure transmitter is 24V.DC. What are the voltages at output levels of 4mA and 20mA respectively?
It is related to the line impedance; generally, it does not exceed 18V. . . Also, why measure voltage when there’s nothing wrong?
It’s fucking confusing me. . . I only know the load resistance is 250Ω; what is the calculated value? . . Hehe, I don’t know; it’s definitely less than 24~
Generally, no one measures this; regardless of the output value, the voltage remains at around 24V. The output impedance can be changed, but the voltage doesn’t change, right?……
The transmitter voltage remains at 24V.DC; the output voltage depends on the load resistance. If the load resistance is 250Ω, and a current of 4–20mA is generated, then the voltage will be 1–5V.
The statement is inaccurate. Does it refer to the voltage across the load resistor, or what voltage exactly? If it is the voltage across the load resistor, then it’s the answer given in point 5.
This post was last edited by HEJIYUER on 2016-3-1 22:03, version 1. The output voltage of 20mA meets the startup voltage of the transmitter (e.g., 10.5V), and the current then varies according to the process parameters. 2. Otherwise, changes in line impedance lead to voltage variations, and Ohm’s law for voltage causes changes in current, which undermines the accuracy of the process parameters. 3. It outputs a constant current; as long as the voltage required to light the lamp is met, it’s fine. @jiaguoyun @Big Horse with Big Bow
Floor 5 is correct; that’s what the question is meant to imply.