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I just tried to measure the current that my circuit draws when powered by 2xAAA batteries (3 V) using my Fluke 87-V. I connected the leads in series with the battery, but I accidentally left the red lead in the voltage socket instead of the mA/uA socket. Is this bad for the multimeter or could I have damaged it somehow?

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The use case that you described would not have damaged your multimeter. It would just have been measuring the "open circuit" voltage of the battery pack as seen through your circuit.

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  • \$\begingroup\$ Yes, of course, I realize now that it's the same as doing a voltage measurement between the +3V battery terminal and the +Vcc rail in my circuit. I'm an idiot. \$\endgroup\$ – David Högberg Oct 28 '13 at 12:59
  • \$\begingroup\$ There may have actually been some cases where meter damage could occur. If your meter had been in one of its special modes such as Ohms or DiodeCheck and the voltage had been high enough there could be damage. But at 3V from two AAA batteries I doubt there would have even been damage in this case. \$\endgroup\$ – Michael Karas Oct 28 '13 at 14:46
  • \$\begingroup\$ Actually, I wasn't entirely right... the meter was in mA/A mode, so it would have tried to measure current and not voltage. Not sure I know exactly what was going on inside the meter. \$\endgroup\$ – David Högberg Oct 28 '13 at 14:50
  • \$\begingroup\$ When the meter is in current measure mode it is really just measuring voltage across a resistor or calibrated wire inside the meter. The extra jack for the current mode is there so that the meter leads can be connected to the ends of the calibrated wire. (Some meters will actually have two of these extra jacks, one for low current to connect to internal resistors and the other for high current that will connect to the calibrated wire). I rather doubt that the regular meter jacks in mA/A mode would experience any problem in the scenario that you had. \$\endgroup\$ – Michael Karas Oct 28 '13 at 16:26

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