# Why is voltage 0 after the last component in a circuit given electron current flows from negative to positive?

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

• Technically it has dropped. It's dropped -9V. Commented Jan 22, 2015 at 5:00
• Because conventional current is a tradition dating from before the discovery of the electron. Commented Jan 22, 2015 at 5:01
• 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. Commented Jan 22, 2015 at 7:49
• @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. Commented Jan 22, 2015 at 8:19
• @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. Commented Jan 22, 2015 at 17:12