Silly question but how would a circuit be set up with Arduino where two different resistors are in series but one has an unknown value (i.e. is just a piece of metal for example) and so how would we be able to find the current through each, collage through each and thus find the value of the resistance of the unknown wire.
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If you know the voltage at either end of the chain, and the voltage at the middle node, as well as the value of one resistance, then you can calculate everything else.– MajenkoCommented Jun 14, 2019 at 12:05
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series resistances have the same current flowing through them ... there is no "current through each" ... it is similar to water flowing through two pipes connected together– jsotolaCommented Jun 14, 2019 at 15:16
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2 Answers
Just a math exercise really.
Say you had a 5V source, two Resistors R1 and R2, and an Arduino. 5V - R1 - R2 - Gnd, with R1/R2 junction feeding the Arduino. R1 is unknown.
Vout is what the Arduino sees. Vout = 5VxR2/(R1+R2) Vout (R1+R2) = 5VxR2 R1+R2 = 5VxR2/Vout R1 = 5VxR2/Vout - R2
Vout = ~ 0.00488mV x Analog In reading. (5V/1024) If you select R2 of say 5K ohm, and you determine Vout to be 3V (analogRead returns 615), then you can calculate R1: (5V x 5000)/3V- 5000 = 3333 ohm.
Checking with Vout = 5VxR2/(R1+R2), 5V x 5000/(3333 + 5000) = 3.000V
While CrossRoads' answer got the math sorted out that is likely not the kind of measurement you are aiming for if the unknown resistor has a rather low resistance (i.e. is just a piece of metal). Assume the same circuit: 5V - R1 - R2 - GND.
Consider R1 to be a strand of copper wire of ribbon cable, AWG 26, i.e. with a diameter of 0.4 mm (0.016 inch) and a cross section of 0.13 mm², and a length of 3 inches (7.62 mm), being that "piece of metal". Its resistance will be 0.001 Ohms. To have any meaningful readings you'd want to have the other resistor (R2) in the same range otherwise the output voltage of the voltage divider will just be close to VCC (or GND if R1 and R2 are reversed). So assuming R2 to be 0.001 Ohms too, the total resistance of both resistors in series is therefore 0.002 Ohms. Connecting these to a 5V power supply (i.e. a voltage source) we find using Ohm's Law a current of 2500 A - which of course an ordinary Arduino wall plug will not provide. The power supply will either limit the current or its output voltage will be reduced or both - either way not a desirable situation in a measurement setup. Another approach of direct measurement is to apply a known current to the unknown resistor and measure the voltage across it. For any reasonable current that voltage will be rather small (i.e. 1 mV at 1 A) and would need to be amplified before being measured with an Arduino's ADC.
In short, measuring low resistances is difficult and might require a dedicated setup such as a Wheatstone bridge. Best answers are likely to be found at Electrical Engineering but need additional input on measurement range and required accuracy.
