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Suppose you have a simple circuit with a voltage source V1 connected to a resistor R1, like this:

schematic

simulate this circuit – Schematic created using CircuitLab

You could connect an ammeter in series, and then the internal resistance of the ammeter would affect the actual current reading, introducing some error. But you could also connect a voltmeter (with a high internal resistance) in parallel across R1, and calculate current by dividing the measured voltage by R1. There would still be some error due to the voltmeter's internal resistance, but which would be more accurate? Or more specifically, under what conditions (i.e. large/small current, R1, V1, etc.) would it more accurate to use the second approach, with a voltmeter instead of an ammeter?

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  • \$\begingroup\$ Are you talking in simulation or with real world devices? \$\endgroup\$ – crj11 Mar 6 '18 at 17:49
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    \$\begingroup\$ if V1 is a constant voltage source, then a voltmeter will not affect the current in the resistor even if the internal resistance of the voltmeter is low. ... the voltage source will simply supply more current for the voltmeter to use. \$\endgroup\$ – jsotola Mar 6 '18 at 17:54
  • \$\begingroup\$ With real world devices. \$\endgroup\$ – goodatthis Mar 6 '18 at 17:54
  • \$\begingroup\$ You can also connect a voltmeter (works well with the analogue moving coil type) in series in the circuit - handy to find faults sometimes... \$\endgroup\$ – Solar Mike Mar 6 '18 at 18:08
  • \$\begingroup\$ Welcome to Heisenberg's world!. Am I being sarcastic? \$\endgroup\$ – Trevor_G Mar 6 '18 at 18:57
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Let's take two examples, one with a high current and low resistance, and one with low current and high resistance. Let's also assume our Ammeter has a resistance of \$1\Omega\$ and our Voltmeter has an impedance of \$1M\Omega\$

schematic

simulate this circuit – Schematic created using CircuitLab

In this circuit, we have a very low impedance source and a low load resistance. This situation is not terribly good for an ammeter, as it's shunt resistance of \$1\Omega\$ is going to change the total resistance to \$11\Omega\$, which is quite a lot of change. However, the Voltmeter has such a high impedance in comparison to the load resistor that it affects it hardly at all. Additionally, since the output impedance of the source is very low, adding another load in parallel will affect the voltage across V1 very little. In this case, assuming the load resistance is accurately known, the voltmeter is the better choice.

schematic

simulate this circuit

In this circuit, both the load and source impedances are high. If we put the Voltmeter in parallel with R1, the \$1M\Omega\$ input impedance is fairly close to R1, and will change it to \$90.9k\Omega\$. However, the ammeter's \$1\Omega\$ resistance will hardly affect the real load resistance, as it's so much lower than \$100k\Omega\$. Additionally, since the impedance of V1 is very high, adding a load in series to it will hardly affect the current it produces. In this case, adding an ammeter in series is the better choice.

As you can see, choosing an instrument with an impedance that is high where the source impedance is low, and an impedance that is low when the source impedance is high are the best choices to minimize the error caused by adding the instrument to the circuit.

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    \$\begingroup\$ +1 TL;DR; you use whichever method introduces the least error based on the meter resistances vs circuit resistances. \$\endgroup\$ – Trevor_G Mar 6 '18 at 18:59
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There are plenty of reasons beyond accuracy, though accuracy does come into it in my first example. Here are a few real-world applications that I've dealt with. First, two from university settings:

  • If there's even the slightest chance current transients or glitches are an issue. An ammeter is too slow to respond but you could swap the voltmeter for an oscilloscope (driving an LED off a digital power supply in constant current mode led to occasional 1mA steps in the current, breaking the experiment).
  • When you want to be able to remove the meter without breaking the circuit or powering down (e.g. setting up multiple laser diode experiments in an educational setting with a simple current source circuit -- measure \$V\$ across a series \$1\Omega\$ resistor built into the drive circuit and you can take the meter away.

Then one (literally) closer to home:

  • In automotive electronics sometimes you have a high current on connecting power, but you want to measure the low current afterwards. You can either (i) set the meter in mA mode, short it, connect the battery, remove the short, or (ii) connect a power resistor in series and measure V across that. The latter is recommended if you only have a finite supply of multimeter fuses. (I only had 200mA and 10A DC ranges available across several meters, the 10A ranges has 100mA precision and I was trying to track down a drain ~40mA)
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