When measuring the I-V characteristic of a metal wire/diode/filament light bulb , what is generally the best circuit set up?

You can

  1. Change the voltage across the component directly using a variable power supply

  2. Use a variable resistor to change the voltage across the component.

  3. Use a potential divider set up to vary the voltage.

In all three cases, you have the voltmeter in parallel with the component and an ammeter in series. You'd have to make adjustments depending on the circuit (e.g. adding extra resistors in series) but as far as I understand, that's the basic setup.

My question is, in 2 and 3 are you not changing the resistance of the whole circuit when you're adjusting the voltage so you're actually changing the current measured by the ammeter indirectly? Is the effect so insignificant that it is ignored? Does this affect the I-V characteristic obtained?

In my view, I think 1 is the best setup for getting the I-V characteristic of a component as you're changing one variable only (which is voltage). Though I'm not sure if I understand the topic correctly.

  • $\begingroup$ If you are concerned about practical implementation (and not theory), you might get a better answer to this question at "electrical engineering SE". $\endgroup$ Commented May 8, 2018 at 21:07
  • 1
    $\begingroup$ @user1583209 I’m asking for verification on what should theoretically happen when you use a variable resistor instead of changing the voltage of the voltage source. I have done this experiment in the lab at school many times and there’s nothing out of the ordinary with my results, but this is only occurring to me now and I can’t find a theoretical explanation as to why this should work in any of my textbooks. $\endgroup$
    – s.xw
    Commented May 8, 2018 at 21:11
  • $\begingroup$ Unless I misunderstand your question (circuit diagrams could help), I don't see any difference between your options. However you have basically two options to place amp meter and voltmeter in the circuit. In one case you will measure the correct voltage in the other the correct current over/through your wire/light bulb/... Errors will be due to nonzero resistance of the ampmeter or non-infinite resistance of the voltmeter. $\endgroup$ Commented May 8, 2018 at 21:30
  • $\begingroup$ @user1583209 In the first circuit, the voltage across the entire circuit is being changed. In the second and third circuit, the voltage across the components are changed but the voltage across the entire circuit is kept constant. I am familiar with the concept that changing the resistance of a variable resistor alters the voltage across the other components in the circuit. But changing the resistance of a variable resistor also changes the resistance of the entire circuit, and hence the current. So is it still possible to obtain fair and accurate readings for I-V characteristic of component? $\endgroup$
    – s.xw
    Commented May 8, 2018 at 21:37

1 Answer 1


As long as the ammeter is in series with the device under test and has negligible resistance, and the volt meter is in parallel with practically infinite resistance, it doesn't matter which method you use. The ammeter measures the current through the device, and the voltmeter measures the voltage drop. That's all you need.

  • $\begingroup$ This sounds absurd, but what if I connect the voltmeter across the variable resistor instead of say a metal wire? I then change the resistance to obtain a range of voltage and current readings. I would surely get similar readings whether I connected the voltmeter across the variable resistor or the metal wire. But that means the variable resistor is ohmic? Where's the fault in my logic? $\endgroup$
    – s.xw
    Commented May 8, 2018 at 20:27
  • $\begingroup$ @s.xw Depending on the rest of your circuit, the voltage dropped by the variable resistor may be nearly constant (equal to the voltage supplied) regardless of the resistance. Not an ohmic response. $\endgroup$
    – BowlOfRed
    Commented May 8, 2018 at 21:35
  • $\begingroup$ I don't understand your question. First, without a circuit diagram, I don't know where you are putting your meters. Second, if I'm measuring the I-V characteristic of a device, why would I put the voltmeter anywhere else but across the device? But if you measure the voltage across the variable resistor you might be able to infer the current through your device, and you might be able to infer the voltage across you device, but it all depends how everything is wired up. A variable resistor ought to be ohmic. Why wouldn't it be? It would be a lousy variable resistor if it weren't. $\endgroup$
    – garyp
    Commented May 9, 2018 at 1:06

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