I am a mechanical engineering student, working on an ohmmeter to test homemade liquid resistors. Something about the way they function means that a normal ohmmeter does not read accurately and gets higher and higher the longer its held up to the terminals. My plan to measure the resistance is to connect the resistor to an AC power supply and measure the voltage drop and current of the circuit to calculate the resistance. My problem is that I am having trouble finding a current sensor accurate enough for my use case. The back of a napkin math I have done says that at 110v AC, with a 100 kOhm resistor (my upper end) there will only be about 0.0011 amps. Is this even possible with an arduino? Any suggestions for sensors would be much appreciated.
The metal you are using to connect to the liquid resistors will likely create a voltage due to electrolysis. Are you using pure platinum contact material? Check the voltage on the connections with a millivolt meter and see what you are getting. Chemistry ! remember it.
Paul
Hello
Check this texasian Hardware.
https://www.digikey.com/en/blog/the-new-ina228-digital-power-monitor.
Have a nice day and enjoy coding in C++.
What might those be?
Chemistry tends to happen when you pass current through a solution.
Can I assume your goal of using an Arduino is to be able to plot data taken over some period of time where a pad an paper with a multimeter is not practical?
First, you wouldn't measure current you would simply measure across the 100k from which you can calculate current.
Do you have any reason to believe the liquid resistor is linear with voltage? Testing with AC could give you some strange results if not. BTW 120VAC has a peak voltage of 120 * sqRoot of 2 = 170 volts at the peak.
JohnRob, as some others have previously stated my resistors have a small capacitance when tested by a normal ohmmeter, so my goal is to test other easier to measure values in order to calculate the value of the resistor. So I cant take your suggestion of calculating current from voltage and the resistance of the large resistor.
That did give me a thought though, Could I put a resistor that is so much smaller that its value while have a minimal impact on the current, but use its know value and the voltage drop across it to find my circuit voltage?
I may have not been clear...
I am under the impression that your "circuit" is:
120VAC ------> 100k fixed resistor ---> liquid resistor ---> 120VAC common (aka ground)
In this case the current through each element is exactly the same.
The current can be calculated by:
I = E1 / R
where E1 = the Voltage across the 100K resistor
Now the "resistance" of your liquid resistor is:
R = E2 / I
Where E2 = the voltage across your liquid resistor
I = the above current.
This will give you the best measurement you can get without specialized equipment.
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