I had this small question today when working with an ammeter. Since an ammeter must be in series of the circuit, I had to constantly break the circuit to take current readings. Whereas, when using the voltmeter, just placing the probes on the device to take an immediate voltage reading works.

Since all devices follow ohm's law, assuming a purely resistive circuit, can't I place a voltmeter and an ohmmeter in parallel to a device to take its current reading?

If yes, why don't we do this often and use ammeters? If no, well why not?

  • \$\begingroup\$ Very few devices follow Ohms law. Resistors do, so place a low value resistor (0.1 Ohm or less) in series with your device and measure the voltage drop across the resistor. Then use Ohms law to calculate the current. This is exactly how ammeters work. \$\endgroup\$ – Steve G Aug 19 '19 at 7:12
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    \$\begingroup\$ @SteveG on the contrary almost everything follows ohms law. What you are confusing is that many things do not have a fixed resistance. But ohms law applies to whatever resistance they have in a particular situation. \$\endgroup\$ – Chris Stratton Aug 19 '19 at 13:11
  • \$\begingroup\$ Can you not simply leave the ammeter connected? \$\endgroup\$ – Solomon Slow Aug 19 '19 at 14:28
  • \$\begingroup\$ A typical, modern ammeter actually measures the voltage drop across a shunt resistor and is calibrated to display the current that must be responsible for the measured voltage. If you can't leave an entire ammeter permanently connected in the circuit, maybe you can leave a shunt resistor permanently connected, then probe that with your voltmeter, and do the math yourself when you need to take a current reading. \$\endgroup\$ – Solomon Slow Aug 19 '19 at 14:31

... can't I place a voltmeter and an ohmmeter in parallel to a device to take its current reading?

Voltmeter across the load is no problem. Ohmmeters need to inject a known current into the system and measure the resultant voltage drop. You can't do that on a live circuit as you will be feeding the supply voltage into the ohmmeter.

A simple solution:


simulate this circuit – Schematic created using CircuitLab

Figure 1. The ammeter can be connected, when required, across the normally closed switch, SW1. Pressing SW1 will cause all the current to flow through AM1 and a reading can be taken. AM1 can be disconnected when SW1 is released.

For currents over an amp or so a clamp-on DC ammeter can be used.


Ohm meters work by injecting a small current into the device and reading the resulting IR drop. So if something else is also forcing current through the device, that would disrupt the ohm meter’s reading. Unfortunately, since you don’t know what the current is then you can’t determine that resistance either. So you can’t solve for the current.

If you have a known resistance inserted in line with your circuit, you can measure the IR drop across that and determine current without breaking the circuit. Many designers will insert a low-value resistance in line with the power supply to make it easier to measure current.


By using Ohm's law you should be able to determine the current flowing through a given resistor, by determining its resistance (before applying any voltage or current, see hacktastical's answer) and the applied voltage.

Keep in mind, that for statistically correct readings you now have to take the accuracy of 2 different values in mind, instead of just one. Depending of the accuracy of the 2 individual devices this can make your calculated current much less accurate than a simple reading in series.

Also don't use this for any non-linear parts. If there's any change of inductive of or capacitive behaviour, this method should be avoided. Use an ammeter in series AND a voltmeter in parallel instead.

  • \$\begingroup\$ Of course this only works if there is a suitable resistor in series with the load or where one can be added. In the case of a capacitive or inductive behavior measuring the voltage across an included (itself non-inductive) resistance is great, because you can measure the voltage as a function of time with a scope, and actually see that behavior in a way that you cannot with an ammeter. \$\endgroup\$ – Chris Stratton Aug 19 '19 at 13:13

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