Perhaps a different way to ask my question is why does voltage drop occur on the negative side of a component and not the positive given what is known about electron current? E.g. in the diagram below voltage would read 9v at node2 and 0v at node1. This would make sense if conventional current were true, but electron current makes me feel like the drop should occur at node 2, not node 1.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Technically it has dropped. It's dropped -9V. \$\endgroup\$ Commented Jan 22, 2015 at 5:00
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    \$\begingroup\$ Because conventional current is a tradition dating from before the discovery of the electron. \$\endgroup\$ Commented Jan 22, 2015 at 5:01
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    \$\begingroup\$ Your circuit is meaningless because the battery/resistor part of the circuit is not directly connected to the ground. It is anyone's guess what the two voltmeters will show. It might be 0V and 9V, but also -4.5V and +4.5V, or 100V and 109V. \$\endgroup\$ Commented Jan 22, 2015 at 7:49
  • \$\begingroup\$ @WoutervanOoijen well it makes perfect sense to me. I mean, probably not what OP meant but that's exactly how you measure a fully diff signal. \$\endgroup\$ Commented Jan 22, 2015 at 8:19
  • \$\begingroup\$ @VladimirCravero, yes I was trying to show the measurement of voltage at the two different nodes with the positive lead making contact with the circuit and the common lead going to ground. \$\endgroup\$
    – Circadian
    Commented Jan 22, 2015 at 17:12

1 Answer 1


Because all instruments still adhere to conventional theory of positive to negative flow instead of the now proven negative to positive flow. In practice, the only thing that matters is that everyone agrees on the same standard, even if the direction is technically "wrong".

  • \$\begingroup\$ It makes sense that the instrument is designed like all other circuits; using the conventional current standard. If electron current was the standard, with ohm's law applied, would node 1 read 9v and node 2 read 0? \$\endgroup\$
    – Circadian
    Commented Jan 22, 2015 at 5:28
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    \$\begingroup\$ @Circadian if it began like that, probably. Remember that voltage is a relative thing - it is always between two points. One could declare Node 2 to be 0V and then Node 1 would then be -9V. Typically, voltages in a circuit are relative to a 0V reference point which is often Earth or the negative terminal of a battery / voltage supply. \$\endgroup\$ Commented Jan 22, 2015 at 6:05
  • \$\begingroup\$ The conventional direction is not wrong per se. You just need to define what is it that you're measuring as moving. Current (gap/absence of electrons) moves from + to - at nearly the speed of light. On the other hand you can walk faster than electrons moving from - to +. \$\endgroup\$
    – slebetman
    Commented Jan 22, 2015 at 8:24

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