Measuring resistance with Ohm's law does not work with sensors because? Does anyone with knowledge about alarms have an explanation

[lopi]

Hello, thanks for accepting me in this group, I come here to share with you all the following situation that I am facing and the truth is I can not find an answer, I have been thinking about it for a long time but nothing. I am trying to calculate the resistance of a closed circuit in series of a PIR sensor, I do know the value of the resistors that the circuit has, because I have done several different tests with resistors of different values, but when doing the calculation according to Ohm's law V: Amps the final value is not correct it is close but it varies I do not see any related pattern here I leave you the way I am operating: I measure the voltage with the red and black cable multimeter, then I measure the amperage of the blue cable in 2 M, I measure the amperage with a multimeter, the blue cable is cut and the multimeter connecting end to end of the blue cable and these are the results "I put the resistors that I know the values" (but they do not match me, I would appreciate it if you told me what I am doing wrong, thank you): V: 13.6 divide by A: .199 = 68.8442 here I have 3k and 2.2k resistors but I think I should get a value similar to 52.000 next ex: 13.6 V .220 A = 61.818 here 2k and 1k resistors. Next ex: 13.6 V .146 A = 91.0958. First of all thanks for taking the time to read this post and thanks for any kind of help.

 

[davelx]

How are you powering the circuit? The source impedance of the supply could be skewing the readings..

A diagram would be helpful

 

[RandomGrinch]

I'm afraid I'm unsure of what you are trying to achieve.

However, as @davelx suggests, you aren't measuring the resistance of the circuit, you are measuring the impedance of the circuit - which is made up of both reactive and resistive components.

A PIR is an active device, apart from the IR detector/s, there may be transistors built into the device, and an IC to interpret the differential pulses received from the detector/s.
This isn't necessarily going to appear as a simple resistance on a circuit.

 

[davelx]

What I was getting at is that all power sources have internal resistance and the available voltage will be split across the internal resistance and the load. This will be the case even if the load is just a simple resistance.

For example, suppose you have a 10V supply and a 1000ohm (1k) load.

You might expect to get a current of 10/1000A, ie 10mA (Ohm's Law). However, this would only be true if the internal resistance of the supply was zero, which it never is.

Suppose that the internal resistance of the supply was also 1k. Then the current flowing would be 10/2000A, or 5mA, and the 10V would split across the internal and external resistances, so you would only measure 5V across the load, even though the supply voltage would measure 10V when open circuit.

Typically, alarm panels have significant internal resistance in the detection loops, which will skew the sort of measurement the OP is trying to make.

It is possible to determine the source resistance. How to do this is left as an exercise for the reader which should aid understanding.