From my understanding

  1. voltage is an electromagnetic force that creates current by exerting a force on an atom that causes that atom to transfer one of its valence electrons to the neighboring atom. The atom that lost a valence electron becomes a positively charged ion and then “steals” a valence electron from the succeeding atom to restore charge balance. This effect propagates throughout the circuit.
  2. Voltage and current are directly related. the larger the voltage/emf, the faster one atom’s valence electron jumps to the next = larger current/electron propagation

Assuming this is in essence all accurate: say we have a simple series circuit with a 5V supply and 2 equal resistors.

The voltage drops throughout the circuit from 5 -> 0, but as we know current is the same across a series circuit

So intuitively, Speaking about series circuits only, is it valid to think of voltage as a “propellant” force or only as a force that “gets the electrons moving” initially and depending on the amount of force, sets the speed at which they “move.”

As an example of this, if we have a person who slingshots a rock in space (space so that the velocity of the rock is constant), voltage would be the force exerted on the rock initially that determines how fast the rock moves through space but after the rock left the slingshots pouch, that force would have no effect on the rock after that point.

Or is voltage more of a “driving force”?

An example of what I mean, if we have a person who’s pushing a boulder on a flat plane, voltage would be the force that’s exerted on the boulder. If the person stopped pushing, the boulder/electrons would stop moving/propagating.

I ask this because voltage drops across the circuit From 5v to 0v but current remains the same, so it seems as if voltage’s only responsibility is to start and set the speed of the current in the beginning of the circuit, but after that current acts independently of that voltage.

  • \$\begingroup\$ Your current analogy with speed is incorrect, as the current is an amount of something (charge). So voltage pushes something against a resistance, and the amount of the something is determined by the push and the resistance. \$\endgroup\$
    – user69795
    Jul 20, 2019 at 23:49

3 Answers 3


It would be incorrect to think of voltage as a propellant, propellants accelerate and most of the time diffuse. Propellants are gasses most of the time.

It is correct to think of voltage as a pressure and electrons as a fluid. Voltage sources are like pumps. Current is like flow. Resistors are like restrictions. That is the best physical analogy of a circuit.

  • \$\begingroup\$ But as voltage decreases across the circuit, current does not slow down. In a water pump analogy, if water pressure dropped across the circuit, wouldn’t the water flow drop across the circuit as well as there would be less pressure driving it forward at each consecutive point of a pressure drop? @VoltageSpike \$\endgroup\$
    – Ietpt123
    Jul 21, 2019 at 1:53
  • \$\begingroup\$ Forgot about resistors, resistors are like restrictions. capacitors are like diaphragm tanks. Wonder what memristors would look like in the fluid world \$\endgroup\$
    – Voltage Spike
    Jul 21, 2019 at 1:57
  • \$\begingroup\$ @lept123 you will not find any mechanical analogs that match up as you want them to because this is not mechanical. You will always find a mismatch if you dig deep enough. At some point, you must discard analogies and treat the thing as it really is: itself, and nothing else. \$\endgroup\$
    – DKNguyen
    Jul 21, 2019 at 2:00

Electromotive Force May be the term you seek



Driving force is more accurate, there's no real inertia to the elections so to keep them moving there always has to be something down the line that's pushing. Like you said there's the valence electrons shifting which drives the electrons at the end of the circuit to all keep moving onward. The more voltage potential, the bigger the obstacles the little electrons can overcome!!

Regarding the relationship between current and voltage, it sounds like you have a chicken-or-the-egg issue with your understanding. Current is a physical property where a circuit will have a defined number of electrons flowing through it per second (Amps = Coulombs/second). Based on the constant current flow through the circuit, each node between resistances will end up at a certain voltage.

For example, if you have two different batteries and attach them to an arbitrary circuit, the higher voltage battery will push harder and drive more electrons through the circuit. Using the current as a starting point, we can then calculate the voltage drop over each resistor and determine the node voltages.

Circuit analysis starts with looking at the circuit from the perspective of the voltage source, and then working out from there. In the circuit below the 12V source has no idea how many resistors there are and it doesn't care, all it sees is that it has to push eletrons through 12\$\Omega\$ of stuff to get the electrons to go through. Using Ohm's Law the current will be \$I=\frac{V}{R}\$ so I = 12/12 = 1A.

Since we know there will be 1A flowing through each resistor we can find that for each resistor a 1A current will cause a 4V drop across each one because of Ohm's Law: \$V=I*R\$ so V = 1*4 = 4V.

Intuitively this makes sense because each resistor is the same so the voltage will be evenly distributed through the circuit, using those same steps you can analyze any resistor circuit as long as you know how to calculate equivalent resistance for parallel and series combinations of resistors.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ But again as I commented on the previous answer, as you said there’s no real inertia to keep the electrons moving so current is highly dependent on voltage, if voltage decreases across the series circuit, how does current remain the same throughout? @KentAltobelli \$\endgroup\$
    – Ietpt123
    Jul 21, 2019 at 2:49
  • \$\begingroup\$ Because at that lower voltage point in the circuit, now there's less resistance to overcome. Analyzing a circuit using Kirchoff's Current Law might help put it into perspective if you've never done that before. It's crucial to look at a circuit as a bunch of nodes, each node can be at a different voltage, but ultimately it's all determined by how big the voltage source is and how big the overall resistance of your circuit is. Then within the circuit it'll all naturally work itself out. \$\endgroup\$ Jul 21, 2019 at 2:51
  • \$\begingroup\$ If we had a series circuit with two equal resistors next to each other, the pressure would dip from 5v before the first to 2.5v between the second and first, so are you saying that the current would be the same as it is before the first resistor as it is in between the two resistors because the wire’s natural resistance at that point would be so low that it compensates for the low 2.5v bringing current back up to equal the current flowing before the first resistor/anywhere else in the circuit? \$\endgroup\$
    – Ietpt123
    Jul 21, 2019 at 3:00
  • \$\begingroup\$ In educational circuit analysis, the wires are assumed to be 0 ohms. So yes the way you're thinking about it is correct and reflects what Ohm's Law says about current flow through two equally sized resistors with the same voltage potentials. Since current is the physical number of elections flowing past a certain point in the circuit (coulombs per second), it's also just a fact of physics that they're the same because there's no other place for the elections to flow except from one end of the circuit to the other! \$\endgroup\$ Jul 21, 2019 at 3:21
  • \$\begingroup\$ So to close the loop on your understanding, 1) voltage is an electromagnetic force that doesn't just act one electron, it's a gradient that acts throughout the whole circuit and 2) yes the larger the voltage the more elections propagate though a given resistance. Does that help? \$\endgroup\$ Jul 21, 2019 at 3:38

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