I'm looking at a circuit (part of a bigger circuit) that consist of a 10 ohm resistor and a red LED.

  • It's running on 2.05V (checked with 2 different multimeters, ranging from from 2.047V to 2.05V).
  • The voltage drop of the LED (again measured with 2 different multimeters on the live circuit) is 1.96V
  • The resistor clocks in at 10.0 ohms exactly (again measured with 2 multimeters).

Now, according to Ohms law, the current flowing through the LED should be

I = V / R = (2.05V - 1.96V) / 10 = 9 mAmps

However, when hooking up my multimeters I get the following readings

  • Multimeter A : 7.30mA
  • Multimeter B : 6.10mA

It took them about 30 seconds to stabilize.

Why am I seeing that much difference ?

Multimeter A is about 4x more expensive than multimeter B, so I guess it will be more accurate, however, the difference between the measured 7.3mA and the calculated 9mA is too big to ignore no ? Or am I missing something ?


2 Answers 2


Your multimeters will have round about 1 or 2 ohms input impedance. They have a small (but not zero) impedance so they can develop a voltage across it which can be measured. In other words they "infer" current by the voltage developed across the small resistor and this small resistor is affecting the current through the LED.

Both meters will have slightly different resistances and hence the different readings.

  • \$\begingroup\$ More like 2-5 ohm ;o) @ddewaele: If you have two multimeters, it should be pretty easy to check the internal resistance of each other meter. \$\endgroup\$
    – jippie
    Commented Nov 6, 2013 at 14:48
  • \$\begingroup\$ @jippie I'm going to check all my meters now LOL \$\endgroup\$
    – Andy aka
    Commented Nov 6, 2013 at 14:53
  • \$\begingroup\$ Is this 1ohm / 2ohm input impedance also called a current shunt ? (like adding a 1ohm resistor in series to measure "current" as voltage) I = V when R=1 ? I now just realise while typing this that that this is probably what the multi-meter does :) \$\endgroup\$
    – ddewaele
    Commented Nov 6, 2013 at 15:03
  • \$\begingroup\$ @ddewaele: Yes. Also see burden voltage (and Fluke note on the subject) \$\endgroup\$ Commented Nov 6, 2013 at 15:07

Andy Aka's answer is correct. I just wanted to add that some meters use different series resistances for difference current measurement ranges. For the Agilent 34401A multimeter I use, the shunt resistance for DC current measurements is specified as 5 Ω when using the 10 mA and 100 mA ranges, but only 0.1 Ω for the 1 A and 3 A ranges. If your meter has the precision (that one displays 6 1/2 digits), you can trade off precision for reduced loading effects, though you may need to disable auto-ranging.


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