In an alternating current, how are frequency, voltage, amperage, and watts related?

For instance, imagining the power as a sine wave, what is amperage if voltage is the amplitude? Is there a better analogy than a sine wave?

EDIT: One of the things I specifically wanted to know is whether frequency and voltage are related? A 60 Hz power signal seems to only be transmitted at 120 or 240 volts - does it have to be a multiple?

  • 2
    $\begingroup$ might be better served on electronics.stackexchange.com $\endgroup$ – JustJeff Sep 26 '11 at 11:27
  • $\begingroup$ Ah, good thought. What should I do, duplicate it? $\endgroup$ – jnm2 Sep 29 '11 at 2:44
  • 1
    $\begingroup$ hit 'flag' like I just did. A mod will come and see these comments and that ought to do it, I'm pretty sure they have an easy button. $\endgroup$ – JustJeff Sep 29 '11 at 2:47
  • 1
    $\begingroup$ Nah, this is on topic here. Electronics.SE is a site for hardware hacking enthusiasts (to quote their FAQ), which means if your question is to be on topic there, you'd better be trying to build something; this question, on the other hand, is about the theory of electrical signal transmission, which definitely falls under the domain of physics. It might also be on topic at Signal Processing, but that doesn't matter since I think this is the right site for it. (For what it's worth, we do have an easy button to migrate questions between sites.) $\endgroup$ – David Z Sep 29 '11 at 4:40
  • 4
    $\begingroup$ 'amperage' is an evil, evil word for 'current'. Although I suppose voltage is also an evil yet alluringly bisyllabic word for 'potential difference'. I guess everyone can call them what they like, but I find 'current','potential difference' and 'power' to be far nicer from a self-description perspective and more helpful in getting an intuitive feel for their interrelation. This said, freqency should definitely be called 'hertzage' :P $\endgroup$ – Richard Terrett Oct 1 '11 at 15:27

He didn't mean 'forget' it. He just meant that it's not really relevant. It's easy to understand the various relationships in DC. When you go to AC they are all the same AT ANY GIVEN MOMENT ON THE WAVE. It all boils down to V=IR and VI=W, RI^2=W, V^2/R=W. With DC it's just a constant. With AC they change with time. If you pick any moment in that time and apply the above, you will have your answer.

  • 1
    $\begingroup$ Oh, and as for using another metaphor, you could offset it by 180 degrees to convert it to cosine, but it will look and work exactly the same. That wave shape (voltage against time) really does accurately describe what happens. $\endgroup$ – Scott Sep 30 '11 at 18:09
  • $\begingroup$ I take it you mean 90 degrees ;) $\endgroup$ – blubberdiblub Oct 5 '15 at 17:30

In the first place, forget frequency, and understand the relationship between voltage V in volts, current I in amperes, and power in watts.

Power in watts at any instant in time is voltage in volts times current in amperes, or $I(t)V(t)$.

If you have an alternating current power source, the voltage and current can change from one instant to the next.

Added: OK, let me try to say a little more about it, in informal terms. On the one hand, suppose you have a big powerful battery, like a 12-volt car battery. It wants to supply direct current (DC) - its frequency is zero (0).

Now you connect an incandescent automobile light bulb to it of, say, 120 watts (a really bright bulb). For that light bulb to dissipate 120 watts of power, at 12 volts, it has to draw 10 amps of current. (By the way, to do that, it has to have resistance of 12v/10a = 1.2 ohms, but that wasn't your question.)

Now, reverse the battery connections to the bulb. It still is as bright as before, still dissipates 120w. The difference is, the voltage across the bulb is the opposite of what it was before, and the current is the opposite also, but when they are multiplied, the power is still positive.

Suppose you reverse the connections 60 times per second, or 120, 400, or a million. How much power does the light bulb dissipate? The same, because there is no time when V times I is not equal to 120w.

Typical wall power is not like that, of course. Rather than instantaneously switching between positive and negative voltage, it swings smoothly with a sinusoidal curve. So the light bulb dissipates maximum power when the voltage is maximum in the positive direction, zero when the voltage is zero, and maximum again when the voltage is maximum in the negative direction. So if the wall frequency is 60hz or 50hz (or 400hz in some airplanes), the power dissipation actually varies between maximum and zero at a rate of twice that. But since the filament takes a longer time to get hot and cool down, it seems to be on steadily. The running average power is constant.

Now, if there are different kinds of load, such as an inductive or capacitive load, or a mixture, you get different power dissipation, because the current is not necessarily in sync with the voltage. But that's another story.

More: In case you're interested in what current and voltage actually are, simple household plumbing is a good analogy. Voltage is like pressure, and current is like water flow. In fact, a piece of metal wire is almost exactly like a water pipe, except it is chock full of electrons that can move freely. They don't actually have to move very fast. It's just that if some are put in at one end, they have to leave the other end at the "same time", because they are not really compressible.

  • $\begingroup$ Um... why would I want to forget part of my question? I want to know. It's actually the reason I'm asking. $\endgroup$ – jnm2 Sep 29 '11 at 13:10
  • $\begingroup$ @jnm2: Because it's an orthogonal issue. It's confusing you. Current, Voltage, and Power are instantaneous properties. You can integrate any of them over time, if you like, to get an average, but that's completely independent. Maybe you could give an example of the kind of thing that prompts your question, and then maybe you could get a more satisfactory answer. $\endgroup$ – Mike Dunlavey Sep 30 '11 at 0:49

Okay, instead of your words "In an electric current...", let's say "In an electrical signal..."

A signal possesses all of the properties you listed above, including frequency (inverse period of oscillation), voltage (energy or amplitude), amperage (charges per second) and wattage (energy per second). The sine wave is a common (and useful) shape for a signal, which means if you were to plot the signal's voltage as a function of time, it would resemble the graph of a sine wave (oscilloscopes are useful because they draw graphs of voltage vs. time). The graph of a sine wave signal can also give you the signal's frequency if you can measure the total time of a single oscillation.

We could plot the current (charges per second) as a function of time as well, and if our voltage graph looks like a sine wave, then our current graph will have a very similar shape. However, the specific relationship between voltage and current is very circuit dependent. In static DC circuits, where the voltage does not change, the current is related by Ohm's Law V=IR. In AC circuits, we have to consider the frequency and phase dependence of V and I.

As a consequence of the complicated V, I relationship, it is a good idea to concentrate on the voltage characteristics of your signal at first, and incorporate the current characteristics as they are needed.

Edit: Power is energy per unit time. If you have a constant source of energy (DC signal) you will have a constant power P = VI. However, in AC signals it is again more complicated, since voltage (energy per time) and power (energy per time) seem to have the same units. At this point it becomes best to describe power by a root mean squared (RMS) calculation, which is interpreted as an average energy per unit time calculation.

  • $\begingroup$ Then is amperage unrelated to frequency? $\endgroup$ – jnm2 Sep 29 '11 at 19:42
  • $\begingroup$ Frequency is a property of any change in current. $\endgroup$ – Greg L Feb 3 '12 at 15:38

Volt (V) is the PRESSURE - potential energy available. Think of a tank full of water on top of a hill. Amp (current flow, I) is the flow of electrons (think water flowing out of the tank). Power (W) is the energy used/ work done when the pressure is being allowed to flow. Resistance (R) determines how current can flow at one time. Think diameter and length of hose connected to the tank and how much water can flow at one time because of it.

V*I=W The amount of pressure available in Volts times the amount of current allowed to flow equals how much work was done.

This is the same for DC ( Direct Current, steady voltage) or AC (Alternating Current, varying voltage over time). At any one instant in time the Voltage will be at a specific point. However sometimes the changes in Voltage and the changes in Amps do not happen at the exact same time. Certain electrical devices are used to intentionally shift one relative to the other. So a changing Voltage at one instant might be at 10V while at THAT instant the Amps might be 0A. So the Power at that instant might be 0W as well. This is called the "phase" or phase angle between the two.

If the voltage varies in a repetitive manner over a given period of time (repetitive pattern of variations) other things can be said about it. Such as how often it repeats. That is called the Frequency (f), how many cycles (...per second, cps, is popular and in electrical measurements is usually used) cps is usually rated in Hertz (Hz). While the Voltage might hit a maximum (Peak) of 10V, it might hit a low of 0V. It might have an AVERAGE of 5V over a given period of time. There are other "averaging" methods as well. Root Means Squared (RMS) is another popular one.

One common repetitive wave is the "Sine wave". It occurs often in nature.

So when you say "imagining the power as a sine wave", They would ALL be sine waves. The V, the I and the W. But they might not all be in sync, have the same phase.

AFA any requirement that some specific relationship needs to exist between a frequency number and voltage number therefore, NOPE!


We know that $V=v\sin \omega t$;

where $V$ is total voltage;

$v$ is maximum voltage; $w=2\pi f$;

$f$ is frequency;
$t$ is variable time;


$V=k \sin(2\pi f)t$; where $k=v$;

$V/k=\sin(2\pi f)t$; $2\pi ft= \arcsin(V/k)$


$f={\arcsin (V/k)}/6.284t$

That's relation between voltage & frequency


Not the answer you're looking for? Browse other questions tagged or ask your own question.