This probably has an obvious answer I'm missing, but searching the web only gave me information on how watts are defined, not why we use them.

Many electric appliances (from things used to build circuits to appliances we use every day) give their power draw in watts. Many other appliances, however, give their draw in amps.

From what I've seen online, watts is calculated by multiplying voltage by current. However, Ohm's law states that \$I = \frac{V}{R}\$, which implies that current already varies linearly with voltage (given constant resistance.) Isn't multiplying current by voltage again redundant?

Similarly to electrical appliances, I've also seen battery capacities defined in both watt-hours and amp-hours, even with batteries that have variable voltages (depending on the charge.)

This doesn't seem to really make sense: if the battery has varying voltages and if amperes don't truly represent power (as I've seen several sources online state as a reason for the usage of watts), why are they used to measure the capacity of batteries that have varying voltages?

  • 2
    \$\begingroup\$ Do you understand what the word dimension means in the context of metrology? \$\endgroup\$
    – Hearth
    Nov 24, 2021 at 2:38
  • 3
    \$\begingroup\$ Why is pressure specified in pounds per square inch rather than pounds? Pretty much same thing with power \$\endgroup\$
    – Maple
    Nov 24, 2021 at 6:22
  • 1
    \$\begingroup\$ I think for even a person who's less knowledgeable, "you can draw up to 1A from this powerbank for 10 hours, or 2A for 5 hours" makes more sense than "you can squeeze 5Wh energy from this power bank". \$\endgroup\$
    – Mitu Raj
    Nov 24, 2021 at 7:35
  • 2
    \$\begingroup\$ @MituRaj Except that a "10Ah power bank" usually has a ~3.8V 10Ah battery, but delivers 5V. So a "10Ah power bank" can't deliver 1A at 5V for 10h, more like 7h. Honestly, Ah makes reasonable sense for a component like a battery. For a product, such as a powerbank, it is borderline useless. \$\endgroup\$
    – marcelm
    Nov 24, 2021 at 11:21
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    \$\begingroup\$ I am afraid how many ordinary people are able to understand 'Watts' or 'Watt hour'. But an average smart phone user does understand what is current and Amperes. \$\endgroup\$
    – Mitu Raj
    Nov 24, 2021 at 13:24

9 Answers 9


Power is specified in watts because a watt is a unit of power. Current is specified in amperes because an ampere is a unit of current. Appliances made for use in homes are designed to run on the AC mains supplied by utilities. In general mains power is generated at a fixed voltage (120 volts in the United States, 240 volts in some other countries). Therefore you can specify the power used by such appliances either directly in watts or in amperes since it is assumed that the latter is for the fixed utility supplied voltage (thus the power can be calculated by multiplying the current by the AC mains voltage). Also most appliances do not appear to the AC mains as fixed resistances so your use of Ohm's law is not relevant.

  • \$\begingroup\$ The US is an interesting edge case because 240V is used for larger loads and a few items can exist in 120V (small) and 240V (large) versions. Then the power in W is essential for comparison \$\endgroup\$
    – Chris H
    Nov 24, 2021 at 15:07

Because 1-amp looks the same whether it is...

A) pushed through a 1000-ohm resistor by a 1000V potential, or,

B) pushed through a 1-ohm resistor by a 1V potential.

Option A requires 1000 watts and will soon vaporize the resistor.

Option B will still get warm but only requires 1 watt.

  • 2
    \$\begingroup\$ "and will soon vaporize the resistor" Depending on the resistor. Open your mind (and your wallet, if you want to buy a 1000W resistor). \$\endgroup\$
    – TimWescott
    Nov 24, 2021 at 4:04
  • \$\begingroup\$ Why would I want use a 1000w resistor when I am looking to emphasize the dramatic difference between 1 amp at 1000v and 1 amp at 1 v. Any good researcher would keep as many variables constant as possible between the two experiments or you hide your result. \$\endgroup\$ Nov 24, 2021 at 4:25
  • \$\begingroup\$ You could use a 1000 watt 1 ohm resistor too! \$\endgroup\$
    – Hearth
    Nov 24, 2021 at 4:38
  • 2
    \$\begingroup\$ Of course, but which version paints the best picture to enlighten the OP? \$\endgroup\$ Nov 24, 2021 at 4:42

Because amps are a unit of current, and watts are a unit of power. They're not the same thing.


An outlet is often just called a power outlet. But if nothing is plugged in, it's not delivering any power. And some I've spoken with were confused by that because they wanted to know what happens to the power if nothing is plugged in. "Does it get recovered or captured and 'put back somehow' if nothing is plugged in?" they asked. "Or does it just spill out and get wasted?"

However, these power outlets do deliver a voltage to each and every outlet. So, to be technical, perhaps the outlet should be called a voltage outlet and not a power outlet, as it always provides a consistent voltage even if there's no power involved as there is no device connected up.

But you have to pick a word and go with it. And in any case, that's what people require (power) when they do plug in a device for use. So "power" it is.

Besides, voltages tend to come in standard lengths. So they don't need to be stated, much of the time. It's either there or it isn't. The value when it is there is just assumed to be known.

I'll switch over to an analogy of sorts.

Voltage is like the distance from a floor to a ceiling in a home. There is a standard height that most folks can just assume. So you don't need to mention a quantity, at all. It's "enough" and that's all you need to know. If you need to look up a specific value, that is easily had.

Current tends to be more like "floor space." It can be anything at all that is appropriate for a use.

You don't need to talk about the height of a room. But people are interested in the area, if you want to tell them. So it would seem odd to hear someone ask, "Could you also tell me how high your ceiling is?" just after having being told that the house is 1300 sq. ft. (or 120 sq. m.).

Sure, the actual living space really is a volume and without the ceiling height you don't know the volume. But you don't care, either. The floor space tells you what you need to know for most uses.

So it may be easier to see why voltage may not be mentioned, at all. (It's assumed to be some standard value.) And why it's enough to mention the current (analogous to floor area.)

In this analogy, power is like the home's volume. That's the actual 3D space required for a decent life, as you can't live glued down flat to a floor (we aren't flounders.) But just saying the area gets across all that's needed for most non-technical discussions.

Floor-space is what you think in, when you are considering some function: kitchen, pantry, bedroom, etc. Yet you really mean volume when thinking that way, because there is an implied working height that you also expect. And there are standards for that. Separately, the functions might be: door and cabinet carpentry, auto repair shop, and so on. And each of these would likely imply a different standard height needed. You need and will pay for the required working volume and the required time it is needed, but you think in terms of the working area and just assume the standards for the height that everyone shares as an assumption for the function. (And you pay for the time you use it.)

Likewise, the voltage for a home is standardized. So there's no need to refer to it. It's a given. The current matters because of the safety breakers and the power matters because of the cost (when considered with time) to operate the device. So while current is a concern ("Will my circuit breaker support this?") and while power is a concern ("Is this freezer more expensive to operate than another freezer?"), the voltage is not a concern. It just exists in the background, so to speak.

That said, like renting space for some function, you will pay for the floor space and the ceiling heights needed (the total volume -- similar to power, here) and also the rental period (time.) In total, the voltage times the current times the time is the energy consumed. And that is what you pay for. So voltage matters in that computation. (If you need 240 VAC instead of 120 VAC and use the same current for the same time, you will pay twice as much for the 240 VAC power usage.) But voltage is just a matter of determining compatibility ("Will this device function properly with this voltage?") What you mostly care about is the current (circuit handling) and the power (cost to operate per unit time of use.)

So it's reasonable for some devices to talk about current. In this way, you can make sure that your circuit, together with other devices that may share the same circuit, will function properly with your breaker. It's also reasonable to talk about power. In this way, you can think about functional efficiency or cost-to-operate. The only use of voltage is for compatibility (a yes-no question of whether or not you can plug device X into socket A.)

But everything, voltage, current, and time goes into computing the dollars spent for some job or task performed.


Power is formally defined as the amount of work done over a period of time. Work is a fundamental concept in physics, and is quantified in units of joules. And the derived quantity of power is canonically expressed in units of joules per second; a watt is by definition 1 joules per second (it is simply defined as such by humans).

Electrical current describes the rate of change of charge over a period of time. Charge is a fundamental quantity in physics and is expressed in coulombs. And the derived quantity of electrical current is expressed in coulombs per second; an amp is by definition 1 coulomb per second (it is simply defined as such by humans).

If you understand and/or accept the fundamental nature of what a joule represents and what a coulomb represents, it is self evident that the derived quantities of power and electrical current are fundamentally different. They are relatable things, but a third concept links them.

That concept is voltage (electromotive force), which is another derived quantity, and which is expressed in terms of joules per coulomb; and again by human definition 1 joule per coulomb is called a volt.

  • \$\begingroup\$ updated thanks @ThePhoton \$\endgroup\$
    – vicatcu
    Nov 25, 2021 at 14:18

Watts are a unit of power, they should be preferably be used. However, for certain appliances such as batteries, amps can make more sense because of the loses inside the battery.

The difference between power, current, voltage has been explained in other aswers, I will only focus on why amp-hours can make sense even if it's not properly a unit of energy.

Let's take an example with a battery that is 12V, 12Wh (1Ah), with an internal resistance of 1Ω

here is a simplified diagram: a perfect source (always 12V) in series with a 1Ω resistor

enter image description here

Let's assume Rload is 11Ω. The output current is $$I = U/(Rg+Rload) = 12V / (1+11)\Omega = 1A$$

the internal perfect source delivers $$ P = U*I = 12V * 1A = 12W $$

the battery will be empty after 1h (12Wh delivered) but the load only received $$ P = Rload*I^2 = 11\Omega * (1A)^2 = 11W $$ $$ E = P*dt = 11W*1h = 11Wh $$

In Ah, the load "received" $$ 1A * 1h = 1Ah $$

So, should the battery be branded as 11Wh ? Not so fast, it depends of the load, let's see with a 1Ω load. This is the maximum power output and can happen when a car starts for example, where a huge amount of power is needed.

now, the current is $$ I = U/(Rg+Rload) = 12V/(1+1)Ω = 6A $$ so the internal source delivers

$$ P = I*U = 6A*12V = 72W$$ so it will last 1/6 h or 10 minutes

during these 10 minutes, the internal source delivered 12Wh, but the load only received $$ P = Rload*I^2 = 1\Omega * (6A)^2 = 36W $$ $$ E = P*dt = 36W * (1/6)h = 6Wh $$ Only half! If counting in Ah however, it is still $$ 6A * (1/6)h = 1Ah $$

Now, the battery delivered only 6Wh, but the 1Ah is still "correct"

In practice, it's even worse than that because the internal resistance isn't constant or even linear, it can vary depending on many factors (current, state of charge, temperature, aging, etc)

TLDR: for a battery, measuring the current output is more accurate than the power because it is depends less of what happens inside the battery, even if it's not a true measure of the available power


it generally changes to what you model the load as, each unit has sort of an implied meaning to it, with a model I mean that you assume something about the load to simplify. Remember that V= I/R is not always true there a lot of conditions that need to be met so you can do that.

I will give you some examples:

Constant impedance : you can think of it as resistance to simplify but it is not exactly the same, the square root of the voltage magnitude controls the power mainly. example Headphones and audio equipment usually mention ohms

Constant current : the voltage will control the power, but we assume the load draws a constant current so from there we can know the power. Transducers and arc lamps behave like this.

Constant power : here you assume the load only has constant power. Some electronic devices behave like this but I got no examples from the top of my head

so each definition implies how it should be modelled as when you work with it, or what conditions are considered constant.

there is nothing wrong with using ohm's law but don't forget the formal definition of electric power


Imagine two rubber balloons, A and B, where B's rubber wall is twice as thick as balloon A.

They are both inflated by injecting air into them, but balloon B's thicker wall will resist inflation more. To inflate them both with 1 litre of air would take much more physical effort for B than A would require, and yet it's the same amount of air in both cases.

In other words, you expend more Joules of energy inflating balloon B than for A, to the same volume, to move the same amount of air from outside the balloons to inside.

To extend this analogy to power, remember that power is the rate of doing work, the number of Joules of "exertion" made each second. The unit is Watts, which means Joules per second. If you intend to inflate both balloons to the same volume of 1L in the same amount of time, obviously you will have to work harder blowing into balloon B than A. In this scenario you are doing more Joules of work per second (power, in Watts) for ballon B than for A.

Conversely, if you don't care how long it takes, then you can make the same effort to inflate both balloons, but for balloon B you will have to make that effort for longer. In this scenario you are delivering the same power to both balloons, but for far longer in the case of balloon B.

The rate of air flow here is analagous to electric current, which is charge flow. Balloon volume would be analogous to total electric charge moved. I hope it's clear by this point that you can't measure power, or work done by knowing only how much air/charge moved, since in all cases it has been 1 litre.

Current tells you how much charged moved, in the same way volume tells you how much air you moved, but it doesn't tell you anything about how much effort the battery or lungs made in order to do the moving. To know that, you also need to know how hard you blew into the balloon, or how hard the battery/mains pushed charge through some device.

For the air analogy, clearly the measure of "how hard" is pressure, and for electricity, that measure is voltage.

If you have to blow twice as hard (twice the pressure) to inflate ballon B to 1L in 10s, as you do to inflate balloon A to 1L in the same time, then you've done twice the amount of work over the same duration, and therefore delivered twice the power. That's why fluid power is proportional to pressure, and electrical power is proportional to voltage.

Now understand that to inflate any balloon to 1L, you can control the flow of air into it, and that rate of flow will determine how long it takes to deliver each Joule of energy. In other words, doubling the air-flow/current will also double the the number of Joule's of work done each second. That's why electrical power is also a function of electric current.

If you double electric current, you double the power, the number of Joules of energy delivered each second to whatever that current is flowing through. If you double the voltage across something, then you also double the power delivered to that thing.

Crucially, what happens if you double both? You quadruple the power. For a fixed electrical resistance, something interesting happens. Let's say you double the voltage across a resistor. Ohm's law tells us what happens to the current:

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

Doubling the voltage across the resistor will also double the current through it, and the combined effect on power is a quadrupling. Similarly, if you double the current through a resistor, you also consequently double the voltage across it. Either case therefore results in a quadrupling of power, a relationship which can be clearly seen in the "square" terms of the two other famous power equations:

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

TLDR; To get to your question, current on its own cannot tell you about energy delivered per second. Voltage alone cannot tell you this either. Only together can you know the power:

$$ P = I \times V $$

Few household appliances have constant electrcal resistance, though an electric kettle comes close. For a kettle, doubling the voltage supply from its usual 110V to 220V would (if the kettle survived) cause the kettle to boil the same amount of water in a quarter of the time, because four times as many Joules of heat energy are delivered to the water each second.

Most electrical appliances are not so simple, and may not be treated as constant resistances. As an example, if you had a multi-voltage TV, rated at 100W, doubling the supply voltage would halve the current, to maintain 100W. That requires some complex switch-mode power supply magic, but it is possible.

Most appliances would just die if you doubled the voltage. Unless they are designed to cope with the sudden doubling or quadrupling of power delivery (which entails some kind of dynamic control of their "apparent resistance"), they usually deal with it by not doing anything any more. Like when you put your back out trying to deliver too many Joules of gravitational potential energy to something in too short a time.


Assuming you know the difference between current and power...

why are they used to measure the capacity of batteries that have varying voltages?

Simple: if the stuff you want to power from the battery draws current that doesn't depend on battery voltage, you'll be interested in battery capacity expressed in Amphours. However if your load uses constant power, then its current will depend on battery voltage, and then battery capacity expressed in Watthours will be more useful. It's not obvious how to convert between the two because battery voltage isn't constant.

An example of a load that current independent of battery voltage would be anything powered by a linear regulator.

An example of a load that draws constant power would be a LED flashlight driven by a switching DC-DC converter. This will regulate a constant output current into the LED, so it has constant output power, and it will draw constant power from the battery (equal to output power + losses) so input current is in inverse proportion to battery voltage.

Many electric appliances (from things used to build circuits to appliances we use every day) give their power draw in watts. Many other appliances, however, give their draw in amps.

If your appliance is a resistor, or if it behaves as a resistor, it makes no difference.

If it doesn't behave as a resistor, for example a switching power supply without power factor correction, then its power factor will be far from 1, and it will draw current in pulses or other weird shapes.

So if you want to use a switch on your wall to control a bunch of LED "transformers" (which are really switching power supplies) you'll be interested in the maximum peak input current of these, so that it doesn't exceed the rating of the switch. In this case, power will be very low, so the switch won't care, but peak current could be a problem.


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