Im currently a senior in highschool. So I know that voltage is the potential difference per unit charge, or from my understanding the work done to move a unit charge from point a to point b without any acceleration. Current is simply the # of Coulombs that pass through a point in the circuit over time. Resistance restricts the coulombs that is able to pass through it. Why is it that higher voltage leads to higher current when the load resistance is constant? Teachers often talk about voltage as potential energy, but isnt the voltage just the work done to move a unit charge between two points. The only way that this makes sense to me is if I think about resistance as the electrons completely stopping or stop at some arbitrary point, and so voltage accelerates these electrons to speed v. And speed v gives you the current.

How does this all tie in with voltage drop? So is it that we say that work is used up? So if we have a current I where the electrons are moving at speed v, do we say that larger resistances "use up" the voltage, while smaller resistances "use less" because the smaller resistances decelerate the electrons less than the larger resistances?

No (water) analogies in your response please. Sorry for the noooby question.

  • \$\begingroup\$ If your teachers know their physics, they should be telling you voltage is another word for electrostatic potential, not potential energy. \$\endgroup\$
    – The Photon
    Dec 4, 2018 at 4:41
  • \$\begingroup\$ Also: Someone tell me if my concept is clear about voltage?. \$\endgroup\$
    – The Photon
    Dec 4, 2018 at 5:06
  • \$\begingroup\$ To all above see my coincidental answer \$\endgroup\$ Dec 4, 2018 at 5:07
  • 1
    \$\begingroup\$ I'm surprised how well you exposed your thinking. Pretty decent. I'll say something and then recommend a good book to read. If you apply one volt difference between two plates in outer space (perfect vacuum, say) and place one Coulomb of electrons at the negative plate surface, they will accelerate towards the positive plate. They will impact that plate with one Joule of kinetic energy, regardless of the plate separation distance. Think about it. The book to read is "Matter & Interactions," 3rd edition (or later), Chabay and Sherwood. \$\endgroup\$
    – jonk
    Dec 4, 2018 at 6:59

2 Answers 2


You are trying to mix sound ideas, voltage being work per charge, and voltage being a potential, with poor ideas that are worse than the water analogy, resistance 'restricting' flow for instance.

The water analogy is pretty darned good, at the level it's intended for, in spite of the fact that water is nothing like electrons. It models well the energy changes around a circuit, and relates pressure to voltage, charge to volume and current to flow rate. That's usually enough to satisfy most folks.

The next model up in terms of 'realism' is the Drude Model. While attractive on the surface, it adds little qualitative descriptive power, and makes such poor predictions for the behaviour of some actual conductors that you have to go full QM to get any correspondence with reality.


The word Potential in the dictionary is accurate for voltage as an adjective. “Having the CAPACITY to develop into something” in this case energy, E.

E = V I * time [watt-seconds or joules].

If the capacitor starts at no charge voltage or ANY voltage difference across the resistor to a Voltage source, then the current immediately rises with the source, Vs.

Then say Vs is held constant after this.

The current drops as the voltage on the cap rises and the voltage across the resistor decays to zero with the current.

When the cap voltage Vc reaches the source Vs the difference across the resistor is zero which represents the current thru it by R=V/I. So you see the fixed resistor indicates BOTH the series current and the shunt voltage across the resistor.


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