I have a material and I want to measure the voltage and current through it over time. Unfortunately I only have an automatic voltmeter and the resistance of the material changes over time. Because I can't measure the current and I don't know the resistance I created a circuit to measure this. See image below.

R1 is the material with variable resistance (disregard the 2kO). I choose a known resistor (R2) and measure the voltage through that. I also know the applied voltage. This way I can figure out R1 as well as the current through the circuit.

The current is I = V2 / R2

The voltage is V1 = Vtot - V2

The resistance is R1 = (Vtot/V2 - 1)*R2 or R1 = V1 / I

My question: will this work? I did the math and it should, however, I have done some tests and it is not working as expected. Should R2 be a value close to R1? Will the voltmeter affect the current path and create a parallel circuit? I need to know the stuff that someone who has some experience with circuits would know.



simulate this circuit – Schematic created using CircuitLab


  • 1
    \$\begingroup\$ The voltmeter does add a parallel path across R2. However, so long as R1 and R2 are small relative to the internal resistance of the voltmeter (on the order of \$M\Omega\$s) you can use this circuit. If the variation in R1 is extremely small however, you will need to setup a precision measurement circuit using op-amps. \$\endgroup\$ Jul 11 '16 at 20:49
  • \$\begingroup\$ Thanks. That actually answers another question I had. By the way, what is op-amp? \$\endgroup\$
    – Gethal
    Jul 12 '16 at 18:39
  • \$\begingroup\$ An Op-Amp is a type of amplifier circuit used in place of discrete transistors as amplifiers. They have several nice properties and can be used in many configurations to perform a multitude of functions, including precision measurements. \$\endgroup\$ Jul 12 '16 at 19:17

I see what you did there with using R2 as a shunt resistor. The reason why it is not working is with your maths or your resistor values. I'm not familiar with the math along with using shunts with such high resistances but it is best to keep R2 as low resistance as possible like 0.1 ohms (yes that small) or less. If your multimeter readings don't go to such small values then you maybe able to get away with 10 ohms. Then you can do the maths you suggested from there. Basically your concept is correct.

Also your voltmeter will affect very slightly to the circuit in the manner of an additional resistor in parallel but it's so small that it shouldn't matter. (It has a resistance of like 10Megaohms generally). If your material has resistances around that region then it's probably going to affect the readings.

  • \$\begingroup\$ Thanks a lot man. I was starting to wonder about that additional path. Quick followup: If R2 is very small compared to the voltmeter then current will idealy flow through R2 and not the voltmeter. Now, if R1 is high (e.g. 3kOhms) will that really matter? I ask because the voltmeter and R2 are further down the path from R1. \$\endgroup\$
    – Gethal
    Jul 12 '16 at 18:44
  • \$\begingroup\$ Firstly current will flow through the voltmeter no matter what resistance R2 is. It's just that since the voltmeter's resistance is so high compared to R2 that by only using R2 in the maths would be accurate enough. Also 3k in R1 is good. \$\endgroup\$
    – Bradman175
    Jul 14 '16 at 8:38

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