I would like to measure the resistance of a resistor in the most basic way by running a current through it and measuring current and voltage. I understand that I have basically two options to place voltmeter and ammeter:

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  1. If I measure the voltage over the ammeter+resistor, I get an error due to the non-zero resistance/voltage drop over the ammeter. So basically the voltage I measure is not the voltage over the resistor but somewhat larger. The measured current is correct though. This method should work well if the measured resistor has a much larger resistance than the ammeter.
  2. If I measure the voltage over the resistor, the measured voltage will be correct, however now the measured current will be larger than the current through the resistor. I understand that if the resistor has a much smaller resistance than the voltmeter this error should be negligible. This method should work well for measuring small resistances.

I have two questions:

  1. If the resistor to be measured has a resistance value that lies in-between those of voltmeter and ammeter, how do I decide which method to use?
  2. If a series resistor (of known resistance) is used, how would I set up volt/ammeter?

Please keep the answer simple as I am not an electrical engineer.

  • \$\begingroup\$ You can try both methods and see what difference there is. Likely, there is none. If there is, though, either your apparatus has awful internal resistance, or the resistance it very close to theirs (either to that of the Ameter, or Vmeter). \$\endgroup\$ May 4 '18 at 7:24

Due to the fact that the characteristics of a voltmeter are far closer to perfect than the ones of a amp meter you'd achieve better results by using your second method. As you noticed correctly an amp meter potentially produces an additional voltage drop of several dozen milivolts where the resistance of a volt meter is that high that the current flowing through it is rather negligible.

  • \$\begingroup\$ As for my second question (with series resistor), I would also use the second method and measure the voltage over both resistors, then use Kirchoff's law to find the voltage drop over the unknown resistor? \$\endgroup\$ May 4 '18 at 7:28
  • \$\begingroup\$ A little bit inconvenient but yes, that would work as well \$\endgroup\$
    – po.pe
    May 4 '18 at 7:32
  • \$\begingroup\$ Why inconvenient? Is there a better way? \$\endgroup\$ May 4 '18 at 7:32
  • \$\begingroup\$ You'd have to know the resistance of your amp meter \$\endgroup\$
    – po.pe
    May 4 '18 at 7:35
  • \$\begingroup\$ Why? I propose to use the second method, where the amp meter is "outside". Instead of the one reistor on the right, I now have two resistors (one unknown, one known). Why would I need to know the resistance of the amp meter in this case? \$\endgroup\$ May 4 '18 at 7:37

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