# How does voltage operate in an electric circuit regarding conservation of energy? [closed]

I hope you all are having a great day. I have some questions regarding the physics of a closed electric circuit on a microscopic level. Specifically, how energy operates and how voltage is "split" across resistive components. My professors always told me something along the lines of "the electrons can sense the resistance and know to split the voltage across all resistive components of the system." It doesn't make sense to me and I'd like some more clarification. Below is my concept of how electricity (charge carriers) work statically and dynamically. It also contains my understandings of specific terms, such as voltage. I'm putting it here in case there is a misconception resulting in me not understanding it.

I believe I am correct to say that voltage is the electric potential energy per charge carrier supplied by an EMF that separates these charge carriers (1 volt = 1 joule of electric potential energy per 1 coulomb of charge). In a static state, the charge carriers contain their electric potential energy. When the electric resistance of a path is adequate enough, the path serves as a medium for the high electric potential charge carriers to flow to a lower electric potential (current). As the charge carriers flow from a high potential to a lower potential, their electric potential energy is converted into a miniscule amount of electric kinetic energy and gain a drift velocity. During their flow, the charge carriers are experiencing a resistive force caused by the electric resistance of the medium material. I believe this resistive force results in the charge carriers to convert some of their kinetic energy into other forms of energy, such as thermal and sound energy (electric power). Although the electric kinetic energy is reduced, the charge density of the medium remains the same, resulting in a lower electric potential difference, but same current. We manipulate this characteristic of energy conversion by providing conductive materials as mediums and making the charge carriers convert kinetic energy into a form of energy we can use for specific duties. For instance, we use this characteristic to provide mechanical energy to motors, or to provide light to poorly lit areas in the world.

The reason I wrote the above paragraph is because I am confused about how voltage works in terms of conservation of energy. The charge carriers have a form of kinetic energy as they flow between potentials of an electric circuit. This electric kinetic energy is converted into other forms as the carriers overcome the resistance of the medium. This implies that the electric kinetic energy going into the medium is higher than the electric kinetic energy leaving the medium, meaning there is an electric potential difference (voltage) between the two points. By conservation of energy, the energy entering the system (entering the medium) must equal the energy leaving the system (leaving the medium). For instance, charge carriers entering a copper wire will have a specific electric kinetic energy; that energy must equal the sum of the kinetic energy of the charge carriers leaving the copper wire and the converted forms of energy.

QUESTIONS:

Why do charge carriers not convert all their electric potential energy into other forms the instance they encounter a resistive force? Why is it that voltage is split evenly across equal resistances, such as a voltage divider? How come the supply voltage is always split across resistive components in a way that the return path is almost zero volts? It all seems so perfect, yet makes no sense to me.

In hindsight, I may just not know how energy conversion operates, but I would still like to know how it works in electric circuits. Any references or advice is greatly appreciated. Hope you all have a wonderful day.

• Electric potential is relative. A particle may be at zero potential relative to one point and at 1000J relative to another, so there is no such thing as "all" potential energy, just energy with respect to some reference. In your resistor example, the electron does lose all of it's potential with respect to the negative terminal of the resistor, but since the resistor's negative terminal will be higher than the negative terminal of the next resistor (Ohms law), it will lose more going through the next resistor. Sep 26 '21 at 15:46
• As for a closed loop having zero net potential, that is a necessary property of a conservative field. If the electric field did not conserve energy then nonzero loops would be possible. Sep 26 '21 at 15:47
• There are two completely different things crossing my mind. One has to do with how charges almost instantly arrange themselves on the conductor surface so as to "direct" current, appropriately. I've already written on this elsewhere, but see the textbook Matter & Interactions by Chabay & Sherwood for detailed descriptions. There is also the concept of potential energy and kinetic energy and the application of the conservation of energy law where $W=U+\frac12mv^2$ is everywhere constant. I've not written about that, here.
– jonk
Sep 27 '21 at 3:41

In a static state, the charge carriers contain their electric potential energy.

I think this is where the misconception comes from: the charge carriers contain .. charge. The energy is embodied in the field.

When the electric resistance of a path is adequate enough, the path serves as a medium for the high electric potential charge carriers to flow to a lower electric potential (current). As the charge carriers flow from a high potential to a lower potential, their electric potential energy is converted into a miniscule amount of electric kinetic energy and gain a drift velocity.

Electric fields behave a little like gravitational fields: if you lift a rock in a gravitational field, that puts a little energy into the rock-field-earth system; if you then drop it, that energy is converted into kinetic energy.

Charge carriers move slowly but persistently for the same reason that pachinko balls keep falling even though they keep hitting pins. They lose their KE, but they can keep gaining energy from the gravitational potential available.

Why is it that voltage is split evenly across equal resistances, such as a voltage divider? How come the supply voltage is always split across resistive components in a way that the return path is almost zero volts?

The short answer is that the field redistributes itself along the current path. The aggregate field is made up of the combination of the field of all the charge carriers, with the field at the terminals of the power source. Remember that voltage numbers are always relative to a reference point, not absolute; there is no "absolute zero voltage".

Why do charge carriers not convert all their electric potential energy into other forms the instance they encounter a resistive force?

I don't know exactly what you mean by this, because all the potential energy lost by a charge carrier on its journey from some point where it has higher potential energy to another point where is has lower potential energy must, as you said, have been "converted" to some other form, like heat or light or magnetic field, by conservation of energy. All of it.

How come the supply voltage is always split across resistive components in a way that the return path is almost zero volts?

The term "return path" is a bit vague to me. Perhaps you mean the point in the circuit that you've labelled as zero volts (say the lowest potential in the divider chain), or the point where the electrons (as opposed to positive charges assumed in the "conventional current" model) have the least potential energy. Either way, the "split", or "sharing" of potential differences across the various elements in the divider chain do not "almost" exactly add up to the voltage source across them all, the sum is exact. Sure, this is a statistical average over many charges, and some individual charges may have behaved in a manner contrary to average behaviour, but on average, the potential energy lost by 1C of charges travelling from one point X of potential $$\V_X\$$ to point Y at $$\V_Y\$$ will be exactly $$\V_X - V_Y\$$. In other words, if a voltage source has a potential difference of exactly 10V, then its electric field is endowing all charges at one end of whatever "thing" it's connected across with 10 Joules per Coulomb more potential energy than charges at the other end.

Consequently, under the influence of this field, and in a typical tendency to move in a direction that minimises their potential energy (as all things do), charges move, and each Coulomb of charges that completes an entire journey through that thing will have imparted 10J of energy to that thing. Not almost, but exactly 10J.

Perhaps, when you imply "approximately" you are referring to the kinetic energy possessed by the charges due to their drift velocity, the "electric current". Where does that energy go? It is certainly present, and is accounted for in the magnetic field associated with that current. There has to be an initial endowment of energy to the charges to accelerate them in the first place, and this energy is embodied in the magnetic field resulting from their average drift velocity. This energy will be returned to the system when the magnetic field collapses.

In practice, when the electric field propelling those charges is removed (voltage source disconnected), their kinetic energy doesn't magically disappear. It is manifest as inductance, causing current to continue to flow (in spite of the absence of an explicitly applied electric field to propel them). Flow continues, and energy continues to be dissipated in the various elements of the circuit as the magnetic field collapses, and the charges decelerate to a halt.

Why is it that voltage is split evenly across equal resistances, such as a voltage divider?

Firstly, voltage is merely a measure of the strength of the electric field at some point, the field that is responsible for endowing those charges with potential energy in the first place, and responsible for propelling the charges towards a place of lower potential energy. By extension, it's a measure of the potential energy possessed by charges at that point.

If you measure +110 volts at point X and +105 volts at Y, you are simply making an observation that each electron at Y (being negatively charged) has $$\110eV - 105eV = 5eV\$$ more potential energy than an electron found at X.

Take for example a conductor that consists of a single length, say 1m, of uniformly resistive material, that is 1Ω per centimetre. You measure 5V of potential difference between the two ends (points X and Y), meaning that each electron that enters the conductor at Y has 5eV more potential energy than it will have when it emerges at X. It's not difficult to see intuitively that at the 50cm half-way point, it must have already "expended" half of its potential energy as heat, and if you measured the potential at that half-way point, you would be measuring its remaining potential energy, which must be 2.5eV less than it had at Y, and 2.5eV more than it will have when it completes the journey to X. A voltmeter will measure that energy as a 2.5V potential difference from the centre point to either end.

Now when you consider this length of conductor to be a set of 100 series-connected lengths of 1cm, each being a resistor of 1Ω, it's clear how the potential at the junction of each pair of resistors must be $$\\frac{5V}{100}\$$ different from the previous junction. This is the principle of operation of the potentiometer used as a potential divider, where the wiper is simply "tapping off" the potential at some arbitrary point along an chain of tiny resistors.

Each unit of resistance traversed by a charge on its way through the medium "costs" the charge some of its potential energy, the same amount for each identical unit of resistance. That amount is related to electrical current, of course, but current is the same along the entire length, by Kirchhoff's Current Law.

You could view the chain of 100 1cm-lengths as two units of 50Ω each. The charge must lose the same amount of potential energy in traversing each unit, and consequently must have lost half of it when it reaches the junction of the two units. Alternatively, consider the chain as two units of 25Ω and 75Ω, one unit having three times the resistance of the other. In that scenario, each charge first traversing the 25Ω section will lose one quarter of its potential energy, and the remaining three-quarters will be lost while traversing the 75Ω section. Measurements of potential difference across these two sections will reflect that ratio of 1:3.

We need only two simple assumptions for current and voltage within a loop.

Neither current or voltage is lost or increased within a loop.

simulate this circuit – Schematic created using CircuitLab

So the summ of all voltages within the loop is zero. V0 + V1 + V2 + V3 = 0

The currents through all resistors and the voltage source are equal. i0 = i1 = i2 = i3

All connections with wires are perfect, the voltage drop is zero. All insulations are perfect, no current flows between unconnected terminals. The behavior of equal resistors is the same.

The only possible solution is: i0 = i1 = i2 = i3 = 10 mA V1 = V2 = V3 = - 1 V

Now to the energy conservation. If we multiply voltage with current to get the power:

V0 + V1 + V2 + V3 = 0 and i0 = i1 = i2 = i3

V0 * i0 + V1 * i0 + V2 * i0 + V3 * i0 = 0

P0 + P1 + P2 + P3 = 0

So all power from the voltage source is consumed by the resistors.

Energy is the product of power and time:

P0 * t + P1 * t + P2 * t + P3 * t = 0

e0 + e1 + e2 + e3 = 0

So no energy is added or lost, energy conservation it true.

• This doesn't answer the "why", which is a far more complicated question. Sep 26 '21 at 18:56