Dear Jaccob Oketcho

I want to share in answering this question but the circuit connections is not fully clear for me by your description.

Would you please make a sketch of the circuit and display it here.

A conceptual answer is that you not short circuit a load but can connect the a resistor with the same value of the coil. In this way nothing will happen and the rest of the circuit continue operating as the initial sate.

Best wishes

Dear Jaccob Oketcho

I want to share in answering this question but the circuit connections is not fully clear for me by your description.

Would you please make a sketch of the circuit and display it here.

A conceptual answer is that you not short circuit a load but can connect the a resistor with the same value of the coil. In this way nothing will happen and the rest of the circuit continue operating as the initial sate.

Best wishes

Abdelhalim abdelnaby Zekry, please see the sketch attached, is it safe to bridge points 13 and 14 and bypass the transmitter?

Dear Jaccob Oketcho ,

Your circuit diagram do not add more information about the nature of the load.

But since this load draws a current of about 4:20 mA, then its resistance is R= 24V/4:20 mA = 6: 1.2 kohm.

So you can connect a resistor with about 3 k ohm to mimic the device.

The power of this resistor p=V^2/3k= 24^2/ 3k= 0.2 watt

You can use 1/2 watt resistor.

You can not connect the 24V at this mA line other wise it will get a fixed voltage with 24 V.

My answer is based on the data you supplied in your question.

Best wishes

Kindly check https://www.dataq.com/blog/data-acquisition/4-20-ma-current-loop-measurements/