hi guys
Does anyone on here have a mathematical example of how the 3 wire RTD when connected to a Wheatstone bridge cancels out the lead wires.

I am new to all this and have just got an understanding of the theory on the Wheatstone bridge and how through changing a resistor we can work out what the RTD resistance is.
When I put in the lead wire resistances (these are just made up for the sake of doing the equation) I can't seem to see how they are cancelled out)

I like to understand things in depth and have watched every you tube clip and looked at almost every web page but can't find the answer I want. Everyone gets to the point of showing the diagram to wire in the RTD. But no one actually does it mathematically. By showing a Wheatstone with resistances before worked out then after the lead resistances are added in. If I could see this, I could see where I'm going wrong.

Please don't tell me that just because one lead is in one fraction of the circuit and one in the other fraction that they are cancelled out. That doesn't make sense to me in the fact that one of those leads would still affect the variable resistor ratio.

have been using Rx=R3x(R2/R1) but I'm opening to the other variations of ohms law.

thank you.

 

Hi Dee,
Welcome to AAC.
This image shows how the 2,3,4 are connected, does that help with your maths?
E

Added link:
https://www.omega.com/en-us/resources/rtd-2-3-4-wire-connections

 

Hi Dee,
Welcome to AAC.
This image shows how the 2,3,4 are connected, does that help with your maths?
E

Added link:
https://www.omega.com/en-us/resources/rtd-2-3-4-wire-connections
View attachment 325323

Thanks for the reply. I had come across that page also. A good write up .

I'll attach my working so someone can tell me where I'm going wrong. Or send through a math example that is right.(I'm struggling to find one)

Basically I take a balanced Wheatstone and then add lead resistance as it would be wired up in the Wheatstone with one leg on the pt100 probe and one leg on another fraction of the circuit.
Using the balanced circuit theory (but knowing my probe is hundred for the sake of the math) I adjust my variable resistance until it matches (with lead added) the resistance of the unknown probe and equals zero volts across circuit.(Balanced)

But if I take that resistance off the variable resistor( working backwards now to check answer but also this is how I imagine it's done in the field)
And use it to try to work out the "unknown resistance" of the probe.
It doesn't give me the resistance as the probe leads are still involved in the equation. Will attach my workings.

 

Really, the solution with the 4 terminal RTD in a complete PERFECT bridge becomes simple because there is no current in the sense wires. That means there is no voltage drop in those wires to be dealt with and compensated for. And the resistance in the one source wire allows calculating the voltage drop in both source wires, based on the assumption of equal resistance in both wires.

The good news is that in a modern setup it is more likely to be a resistance measuring circuit than a true bridge circuit, with a constant regulated current thru the RTD.