Can someone please explain to me the concept that there is lesser current flow for a higher applied voltage, to deliver the same amount of power?

I want something close to a physical approach explanation.

I already looked up similar questions and I am not satisfied with explanations relating Ohm's law (V=IR) or P = IV (purely mathematical reasoning). Also please do not use the water pipe analogy for the sake of mankind.

Here is my thought process:

  1. Power is work done per unit time (joules/sec)
  2. Voltage is joules per coulomb(work per unit charge), one joule of work is present for a charge amounting to 1 coulomb.
  3. Voltage as an electric potential, means that there is 1 joule of work stored as electric potential energy, for 1 unit charge of coulomb.
  4. The higher the voltage (electric potential), means larger joule of work for each coulomb.
  5. For a higher power demand means higher work per time (joules per second). A higher voltage can provide a higher amount of work (joules) for each charge (coulomb).
  6. Ampere is coulombs per second.
  7. For a certain amount of coulombs per second flow (current), a higher voltage can output a higher amount of work for each coulomb, compared to that of a lower voltage.
  8. Since power is the amount of work done over a period of time (joules/sec), the higher voltage delivers a higher amount of work per second, per coulomb.
  9. If we used a lower voltage to output the same power, the Coulombs per second have to increase (higher current/amperage) because the lower voltage has lower joules for each unit coulomb, leading to lower Joules per second, per by the unit charge coulomb.

I am a bit confused with my point #2 being that work (joule) is carried through a unit charge (coulomb,) since point #2 must be established for the proceeding points to make sense. Is this model or approximation of energy transfer correct?

I would like that someone correct all the points I have made, or point out mistakes in my thought process.


3 Answers 3


Your thought process leads to a correct statement in (9) - when providing a constant power, if the voltage decreases, then the current must increase. This is a correct statement, and your original goal (if the voltage increases, the current must decrease) is also a correct statement.

A simple deduction is as follows:

  • We require a certain number of joules to be delivered to the load each second.
  • If the voltage is increased, then each coulomb of electrons provides more energy as it traverses the potential difference associated with that voltage (i.e. across the load).
  • With the increased voltage, we need less coulombs of electrons delivered in each second to provide the requisite number of joules.
  • Less coulombs per second means less current.

However, while these hand deductions are useful and illustrative, they're a bit verbose and unwieldy to use throughout every calculation. P = IV is not "purely" mathematical reasoning, and it is actually a very strong and accurate representation of the physical relationships between power, voltage, current, charge, and energy that you point out.

We can transform your statements (1) and (3) to the equations \$P = E / t\$ and \$V = E / q\$ (with E representing energy, P representing power, and q representing charge to match the notation used in my part of the world). We can also rearrange (6) to \$I = q / s\$.

Finally, we can reach the statement \$P = V * I\$ (which you were unsatisfied with as "purely mathematical") as an algebraic rearrangement of mathematically-stated physical statements.

This should come as no surprise: After all, the point of mathematics is to model real-world phenomena in a rigorous way, and with careful use of notation and dimensional analysis, mathematics is a perfectly suitable model here that restates physical statements, much like the ones listed in your question and at the top of my answer, just in a more compact, rigorous, and precise manner.

I am a bit confused with my point #2 being that work(Joule) is carried through a unit charge(Coulomb), since point #2 must be established for the proceeding points to make sense. Is this model or approximation of energy transfer correct?

If one coulomb of charge is moved through an electric field and experiences a total potential change of one volt (i.e. one joule/coulomb), then one joule of work has been done. The work of one joule isn't inherent to that coulomb, but rather it is a consequence of both the charge and the path it takes through the circuit.

As an aside: I will agree that V=IR is, in some sense, a mathematical formula that models empirical behavior and is not clearly physically rigorous. People did practical experiments with materials, found that many of them did have voltage proportional to current, and proposed V=IR as an empirical model that was useful to model a bunch of materials used in practice. There are derivations of it that use more modern theory like linear response theory or quantum mechanics, but ultimately it is much less "physically justified" than P = IV (which I justified above). There are plenty of materials that don't follow Ohm's Law, but P = IV holds perfectly consistently when current passes through them.

Thankfully, we don't actually need to use Ohm's Law to prove the statement you wanted. That statement holds (and P = IV likewise holds) regardless of whether the load is a resistor, something else that follows Ohm's Law, or some non-ohmic, non-linear material.

  • \$\begingroup\$ (1/2)Thank you for your insightful response. >If the voltage is increased, then each coulomb of electrons provides more energy. Isn't it wrongly assumed that "each coulomb of electrons" carries a variable amount of energy due to an increase in voltage? What I am saying here is, how the coulomb of electrons store this energy? If we look at voltage here as the measure of stored energy of Joules much like how we attach this "coulomb of electrons" to a spring, higher spring constant k means that there is much available potential energy stored to move the charge between two potential points.... \$\endgroup\$
    – rthony
    Commented Apr 26, 2023 at 15:40
  • \$\begingroup\$ @rthony Each coulomb of electrons provides more energy as it traverses the potential drop associated with the voltage you applied to your load. \$\endgroup\$
    – nanofarad
    Commented Apr 26, 2023 at 15:43
  • \$\begingroup\$ (2/2)The potential energy stored in moving the charge to and from a defined distance is stored in the spring, if granted that this spring analogy is true, how then is the potential energy stored in these "coulombs of electrons" as you stated in your point which says """If the voltage is increased, then each coulomb of electrons provides more energy.""" \$\endgroup\$
    – rthony
    Commented Apr 26, 2023 at 15:44
  • 1
    \$\begingroup\$ I don't state that the energy is stored in those coulombs of electrons. The energy is a consequence of both the electrons and the fields/potential gradients that they cross at the load (and consequently, also at the generator which must do work upon those electrons). Rather than using a spring analogy, a gravitational energy analogy may be more useful - a kilogram mass lifted higher above the floor can provide more energy when it is dropped or lowered. The energy isn't inherent to the kilogram mass, but a consequence of the mass and the Earth's gravitational field. \$\endgroup\$
    – nanofarad
    Commented Apr 26, 2023 at 15:46

Rather than wade through your long argument, trying to figure out what's exactly right, wrong or useful, I'm going to cut the Gordian Knot with a simple scenario that shows you that to deliver the same power, a higher voltage needs less current. You can then use that to inform your intuition.

Consider a 1 V battery, driving a resistance of 1 Ω. It's delivering a power of 1 watt, power = V x I.

Now consider a second identical battery and load.

  1. Put the two batteries and loads in parallel.

There will be 1 V across the composite parallel load, which will be consuming 2 A, and so dissipating a total of 2 W.

  1. Rearrange the batteries and loads into series.

There will now be 2 V across the composite series load, which will be consuming 1 A, and so dissipating 2 W.


simulate this circuit – Schematic created using CircuitLab

The power from each battery, and into each resistor, has not changed with the rearrangement. However, the current and voltage measured through and across the composite batteries and loads has.

Now go figure.


The water analogy isn't going to help much here anyway. It's good for some scenarios, but not this one. Your bullet points are all sound, and there's nothing to correct. You seem to have got it. The best I can do is present the topic in a slightly different way:

  1. Voltage is a measure of the potential energy possessed by a charge. Since everything's relative, a "voltage" is considered to be the difference in potential energy between charges at one point in a system, and charges at another point.

  2. Potential energy is the energy that the charge could donate to something, if it were permitted to, in the same way that if you hold a heavy, expensive clock above ground, it has gravitational potential energy, the potential to smash itself to pieces. It won't, though, unless it is permitted to fall and hit the ground.

  3. Current is the measure of the number of charges passing some point in a conductor during a certain interval of time. This can also be interpreted to be the number of charges covering a certain distance in that same interval. It's like having a single lane of cars travelling at 100mph along a road being equivalent to two lanes of cars travelling at 50mph. In both cases, the same number of cars pass by an observer in some given interval of time, and the same number of cars make a complete journey of 1 mile in the same interval.

  4. When I say "distance" above, I mean "potential difference". The actual distance a charge travels is irrelevant, what's important is the potential energy difference that it traverses.

  5. "Work done" is the energy that a charge gains or loses on its journey between two points of different potential. If a charge moves from a point in a system with a potential of 10V (10 units of potential energy per charge), and moves to a another point at 2V, it will have lost 8 units of potential energy, and that energy will have been donated to whatever it encountered on the way, to make heat, light or motion. The charge has done 8 units of work. When a charges moves from a place of low potential energy to high, it must have received energy from the environment. The environment must have done work on the charge. This is what occurs when you charge a battery.

So, here's the answer:

If you double the voltage across some conductor, you double the difference in potential energy between charges at each end. You double the work each charge can do as it makes the journey from one end to the other.

If you then also halve the current, you halve the number of charges that make that journey from end to end each second.

The net effect is that the same work is done on the environment per second, which is another way of saying the same power is dissipated in the conductor.

Perhaps your confusion stems from the fact for a given electrical resistance, if you double the voltage across, you also double the current through it, by Ohm's law. This points to the impossibility of the above scenario for a fixed resistance.

If you want to double the voltage, and halve the current, you have to increase resistance by a factor of 4. There's no other way.

As a side note, doubling the voltage doubles the potential energy possessed by each charge, which it will donate to the environment during on its journey. If you keep resistance constant, then current will also increase by a factor of 2, doubling the number of charges making the journey each second.

That's an increase of power by a factor of \$2^2=4\$. That's what gives rise the the squaring in the famous power laws:

$$\begin{aligned} P &= I^2R \\ \\ P &= \frac{V^2}{R} \end{aligned}$$

  • \$\begingroup\$ Thank you for your response. Regarding point number 2: Gravitational potential energy being similar to that of electrical potential energy: The former is a product of mass, g force, and height. How would you individually relate these parameters to the latter? Could we say, mass is the coulomb, pertaining to the amount of matter or charge, while the product of g force and height relates to the work stored in moving the coulomb between two relative potential points? If this is true, I am curious how a coulomb of electrons stores this energy. \$\endgroup\$
    – rthony
    Commented Apr 26, 2023 at 15:58
  • \$\begingroup\$ @rthony I think I agree; the work done/energy stored in both cases is force × distance, force being proportional on the mass/charge. It takes force × distance = work to move a mass/charge in a direction opposing where it would naturally "fall" in a gravity/electric field (like lifting a ball or charging a battery), and if left to "fall", a mass/charge will do force × distance work on the environment (like a clock smashing or an LED glowing)....continued... \$\endgroup\$ Commented Apr 26, 2023 at 23:46
  • \$\begingroup\$ If you have a system in which charges are uniformly distributed, there would be no net force on any charge to move it anywhere, all charges have lowest potential energy. If you somehow moved a bunch of electrons together in the same direction, this would disturb that equilibrium, creating an electric field between the more "net-positive" side and the more "net-negative" side, a field in which all charges would experience a force trying to restore that equilibrium...continued.... \$\endgroup\$ Commented Apr 26, 2023 at 23:57
  • \$\begingroup\$ It's like deforming a spring; it wants to spring back when you remove the force deforming it. The energy "stored" in the spring in this way is actually electric too - by moving charges so that their equilibrium positions are disturbed, energy is imparted to them. When the force that causes that disturbance is removed, they all want to return back to their state of lowest potential energy. It's all about charge distribution, forces between charges in the system. \$\endgroup\$ Commented Apr 27, 2023 at 0:01

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