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As far as I understand, voltage is a difference of potential energy between two points, and that current flows from the negative to the positive pole of a battery in a circuit.
Thus, according to these premises, voltage would decrease from negative to positive as the potential energy is used to move the electrons.
In a multimeter, does the same logic apply, and is the energy at the red lead subtracted from the black? Or is it the other way around, and why?
As far as I understand, voltage is a difference of potential energy
Just potential. Or voltage. You can express the voltage as 'potential energy per unit charge', but that's a bit of a mouthful.
between 2 points
Exactly, I'm glad you have that
and that current flows from negetive to positive poles of a battery in a circuit.
Conventional current flows from positive to negative, by convention. Current is made up of the movement of charge carriers. Those charge carriers can be anything mobile that carries charge, so positive and negative ions in an electrolyte, electrons and holes in a semiconductor, electrons in a wire, electrons and positive ions in a plasma. Positive carriers move positive to negative, negative carriers move negative to positive. As it happens, in metal wires, electrons are the only charge carrier, and they move negative to positive. But they are electrons, not 'current', which we always reserve for 'conventional current'.
Because they are negatively charged (by convention), the direction of conventional current is opposite to the flow of electrons that it consists of in wires. Be kind to the pioneers, they had to choose a direction for current ages before the electronic/nuclear theory of matter.
This may seem like an awful shame, as people first meet, and often only meet, electrons as charge carriers, but any time we have to deal with carriers of both signs, we need to consider their movements separately, so we would always have some that were negative.
Thus, according to these premises, voltage would decrease from negetive to positive as the potential energy is used to move the electrons.
No, the positive pole is defined to have a higher potential, a higher voltage than the negative. Whether it does, or the opposite, is down to convention. You're free to use whatever convention you like in your own work, but 100% of the scientific and engineering world has chosen the '+ve is higher convention' (for better or worse), and if you want to be able to communicate with them, you should too.
Now my question is, in a multimeter, does the same logic apply, and is the energy at the red lead subtracted from the black? Or is it the other way around and why?
A multimeter reads the voltage difference between the leads, and reads positive for the red lead being at a higher potential, a higher voltage, than the black. We usually express this as the voltage of the red lead 'with respect to the black'.
We often simplify this long-winded way of measuring voltages by choosing a terminal to be our '0 V' reference, connecting the black lead to that, and measuring all other voltages with the red lead. Which is fine, as long as we always remember that we are really dealing with voltage differences.
They show "red-lead" potential minus "black-lead" potential.
it subtracts the black from the red, if red is positive and black is the other.if your voltage reading comes back negative, the black one is higher. depending in your application, this could be correct or you might have them plugged in backwards.
but your meter takes the difference between the two leads already. so you dont really have to with one reading by itself. if you need to get the difference between two nodes for example, measure each and then subtract one from the other depending on what youre trying to do
