Electronic – Possibly Incorrect Current Reading

currentcurrent measurementmultimeter

I'm not entirely sure if I have done the following equation correct or not, but based on the calculations, my multimeter result seems to be incorrect. I have a very simple circuit, containing a 9V battery, an LED (with 3.2 V forward voltage at 24 mA), and a 270 ohm resistor (as shown below).

schematic

simulate this circuit – Schematic created using CircuitLab

My question is, when I do the calculations to determine what the current should be in the circuit, I do (voltage / resistance = I), so I get (I = 5.8V / 270 = 0.0214 Amps). This is 21 mA, approximately. When I insert my meter into the circuit, (inserting the meter between + and the LED, breaking the circuit and properly measuring amperage) I obtain a reading of about 2.1 mA. Did I do the calculation wrong?

I am using a Craftsman 82344 autoranging multimeter. Also, when I insert the multimeter into the circuit, the LED no longer illuminates as it should (at least according to the articles I have studied). Any help would be appreciated, if any more details are needed let me know.

Best Answer

Since your circuit has a known resistance in it, you can avoid the meter burden voltage issue by calculating the current from a measurement of the voltage drop across the resistor measured with your meter in voltmeter mode, rather than by inserting the meter into the circuit in ammeter mode.

Of course, if your resistor doesn't have the value you think it does (manufacturing tolerance or human error in reading the markings) that determination will be invalid, but you can remove the battery and use your meter in ohm mode on the unpowered circuit to measure the resistor.