No, no amplification. The voltage drop across the combination of the transmitter and the resistor at the input card is determined by the DC power supply. The transmitter will regulate its own apparent resistance to deliver the required current for the signal it is transmitting.
This is where I get lost. Right now if I take a meter (see below)
TB3 1-2 = 24vdc
TB3 1 -TB4 1 or 2 = 24vdc
But @pturmel how does the below turn into 1-5 if I clearly have 0 volts.
TB3 2 - TB4 1 or 2 = 0vdc
TB4 1-2 = 0vdc
so the input is getting 0 volts as it stands. We add resistance which doesn’t drop voltage but “adds” it? i.e. I can’t drop from 0 to get 1-5vdc…?
No, the transmitter changes its voltage drop to deliver the required current. If the downstream resistance is zero, its own voltage drop will consume the entire power supply voltage. After you add some downstream resistance, the drop will be split between the transmitter and the load.
By adding the resister, what you are doing is creating a voltage divider. You have to go way back to you basic electronics classes to remember the example problems the textbook gave you! Here is a link to a Wikipedia article to refresh your memory: