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Basically the subject. Why would companies bill for wattage instead of amperes?

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    \$\begingroup\$ @SolarMike Are you sure? IIRC, here home lines are billed per kWh and larger companies are billed per kWh with restrictions on their power factor. Better power factor = better price, so indirectly billed on kVARh but not exactly. \$\endgroup\$ – Mast Apr 6 at 8:10
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    \$\begingroup\$ @Mast and the rates are variable below X is one rate, above another and then those rates can change 2 or 3 times per day - bills can be very complicated... \$\endgroup\$ – Solar Mike Apr 6 at 8:12
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    \$\begingroup\$ Caravan sites and marinas are a common example of fixed-supply billing - in the UK stringent regulations and type-approval apply to meters used for energy resale, and it's a pain to have to read every meter every day/week, so a fixed charge is used. \$\endgroup\$ – Owain Apr 6 at 12:10
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    \$\begingroup\$ How much do you pay for a Watt then? \$\endgroup\$ – Dmitry Grigoryev Apr 6 at 13:03
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    \$\begingroup\$ Compare for cars: "why do you buy fuel by the gallon/litre rather than distance?" \$\endgroup\$ – Viktor Mellgren Apr 7 at 13:34

12 Answers 12

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Why would companies bill for wattage instead of amperes?

Because amperes don't tell the full story about energy transfer from a source to a load. If you supplied a load that took 100 amperes at 1 volt, the power consumption (joules of energy per second) is 100 watts. If a different load took 100 amperes at 100 volts, the energy transfer per second is 10,000 watts.

If you only billed in amperes you bill both customers the same.

It's all about energy and power so, you calculate power delivered to the load and that accumulation of power with time (energy) is what you are billed on. To calculate power it is amperes x volts.

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    \$\begingroup\$ Do you know any power company that delivers power at 1 V? \$\endgroup\$ – The Photon Apr 5 at 18:11
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    \$\begingroup\$ USA 120V, most Europe and India 220V... so there are variations @ThePhoton \$\endgroup\$ – beccaboo Apr 5 at 18:19
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    \$\begingroup\$ France : before ~1974 it was 110V, and after 220V \$\endgroup\$ – andre314 Apr 5 at 18:21
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    \$\begingroup\$ Who asked the question has a very little amount of knowledge on electricity. So, if you guys tell him about power factor, higher charge for industrial load, etc then he might not understand anything at all. And 1 volt is just an example. Using 1 and 100 makes a statement easy to understand. \$\endgroup\$ – Sadat Rafi Apr 5 at 18:49
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    \$\begingroup\$ Actually companies does not charge by the watt (that is a power unit) but by the Wh (that is an energy). 1W = 1J/s, 1Wh = 3600J. \$\endgroup\$ – DDS Apr 7 at 14:58
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They don’t bill in watts (power) either. They bill in watt-hours, that is, energy consumed. (kilowatt-hours typically.)

Let’s break this down a bit. Current alone doesn’t tell you power. You also need to know the voltage, as power is voltage * current. Then, you tally power over time to figure energy.

Now you could estimate power from current if you make an assumption about the voltage, and in the early days of electricity they did exactly that using ‘coulomb counters’, that is, they measured and tallied the current delivered over time and billed based on the tally.

This current-only method proved to be inaccurate because of line fluctuations, so for this and other reasons (notably, the adoption of AC power) they developed the motor-type meter, and later, the more familiar spinning-disc induction-type watthour meter. These meters also take voltage into account by design.

More about meter development here: https://www.smart-energy.com/features-analysis/the-history-of-the-electricity-meter/

And because it matters sometimes, a discussion of real vs apparent power and how utilities deal with it: https://www.electronicdesign.com/power-management/article/21806945/how-does-power-factor-correction-impact-your-utility-bill

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  • \$\begingroup\$ P = V x I if they are directly in phase. Where is cosine? \$\endgroup\$ – relayman357 Apr 5 at 18:14
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    \$\begingroup\$ More about apparent power. Utilities will add a power-factor penalty for users that have low power factors as their reactive loads still tax the grid despite not dissipating real power. This doesn’t apply to household users who don’t tend to have large reactive loads. \$\endgroup\$ – hacktastical Apr 5 at 18:58
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    \$\begingroup\$ That all said, if you marked me down because of this phase angle consideration, shame on you. It’s beyond the scope of the OPs question, and it doesn’t even matter as typical meters measure real power anyway. \$\endgroup\$ – hacktastical Apr 5 at 19:07
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    \$\begingroup\$ @relayman357 - if you integrate the instantaneous product of current and voltage the phase is automatically accounted for as well as any distortion in the signal. \$\endgroup\$ – Kevin White Apr 5 at 19:44
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    \$\begingroup\$ Hi Kevin, agree. Not just V x I as typical meter readings (which are typical RMS volts and amps). V times I at each point in time from waveform sample data results in power waveform. If phase angle between V and I is 90 (pure sinusoids) then P = 0. \$\endgroup\$ – relayman357 Apr 5 at 19:50
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I think because they care about energy consumption, energy is what costs money and resources to generate, also you can have different voltages when using 2 and 3 phases which would make the charge measurement by itself useless

-- edit, and yes, they measure in watt*hours for obvious reasons

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One reason why they bill Watt-hours is because that's what is directly measured by the classic, electro-mechanical service meters that were used all over the world for more than a century. Many of those meters still are in service today.

Modern electronic meters measure voltage and current separately. They could be programmed to total up the Amp-hours, but it makes sense for them to calculate Watt-hours, for compatibility with the older meters.

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Most utilities both bill energy [kWh] and demand [kVA] as well as a fixed monthly charge for large customers. There are various other costs that can be billed.

The demand component is a way to bill the current, \$ [\textrm{A}] = \frac{[\textrm{kVA}]}{[\textrm{V}]}\$ because the voltage is mostly constant.

The current is integrated over the integration period to get an average.

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Companies bill for the Watt.hours (Wh) product because that's the thing that makes up the irreducible cost of supplying the energy. If you use more Wh, then the supply company has to burn more fuel. There's a direct proportional relationship between Wh supplied and kg of fuel burned.

Within that Wh product, or Volt.Amps.hours product, it's possible to make any of those terms big or small without impacting how much fuel is used. You can trade off Volts against Amps with a transformer. You can trade off power against time with storage.

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kWh provides part of the basis for a method of billing which reasonably reflects the actual cost to many utilities and is reasonably fair to the customer in a majority of situations. There are, however, situations in some parts of the world where the current capacity of the system is limited and the grid connection limited or non-existent and there is only limited capacity to modulate the delivery of power -- e.g. hydro-electric schemes in remote areas with limited or non-existent grid connection. In that sort of situation even domestic premises are billed according to maximum current draw and there's a large ammeter on the kitchen wall with a red line indicating the maximum permitted. Go over that and the financial penalties are severe.

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Actually companies charge by the Wh (Energy), not by the W ([instantaneous] Power).

1W measures the "rate" at wich you use power. 1kW heater uses 1000J/s (1000J every second)

1Wh measures the energy. 1kW electric heater running for one hour uses 1kWh = 3,6MJ Energy.

A 100W light bulb would use the same kWh [3,6MJ] in 10 hours.

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Many devices spend part of each cycle taking more energy from the utility than they're going to use, but then give the unused portion of the energy back during another part of the same cycle. If a meter were to merely measure RMS current rather than power, it would not only fail to credit devices for the energy that's given back to the grid, but it would actually bill users for that energy as though they were consuming it.

Note that utilities do often bill big customers for current as well as for power, because the utility will lose a fraction of the power that flows through its wires. A utility might have to generate 1,010 joules in order for 1,000 to reach the customer. The fact that there are 10 joules the customer isn't directly paying for wouldn't be a problem if the customer would pay for 1,000, but if the customer gives back 990 joules, and 980 reach the utility, the utility would end up with 30 less joules of energy than it started with, while only getting paid for 10. For small residential customers, this isn't a big issue, but for large industrial customers it can be huge.

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  • \$\begingroup\$ There is no difference in measuring negative current or negative power. Both would be possible, and since mains is pretty stable it is just a linear factor difference. At least as long as rms current and apparent power. \$\endgroup\$ – eckes Apr 7 at 18:44
  • \$\begingroup\$ @eckes: Whether power is positive or negative depends upon whether current and voltage are positive and negative simultaneously, or whether voltage is positive when current is negative and vice versa. Sampling current alone would not convey that information. \$\endgroup\$ – supercat Apr 7 at 18:58
  • \$\begingroup\$ The same mechanism used to detect if power is consumed or provided can be used to determine if it's a consumer or provider current. \$\endgroup\$ – eckes Apr 7 at 19:09
  • \$\begingroup\$ @eckes: Suppose a device connected to a 60Hz mains causes 1 amp to flow from the hot wire to the neutral for one millisecond every 16.6667 milliseconds (1/60 of a second). If the hot wire happens to be at +170V at that time, the device will take 0.17 joules from the utility. If it happens to be at -170V, the device will push 0.17 joules back to the utility. If it's at e.g. +30V, the device will take 0.03 joules from the utility. If it's at essentially 0V, the device won't take nor deliver any meaningful amount of energy. The amount of energy transferred by moving... \$\endgroup\$ – supercat Apr 8 at 16:18
  • \$\begingroup\$ ...a certain amount of current depends upon precisely when within the AC cycle the transfer takes place. One could perhaps build a meter that measured current and the phase of the line voltage without measuring the magnitude of line voltage, but it's simpler to just measure power. \$\endgroup\$ – supercat Apr 8 at 16:19
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Consider two customers who are using electric heat and who both need precisely the same amount of energy to heat their homes. Imagine this is the United States and the official voltage is 120V.

Say one customer is a bit further from the transformer though and they are getting 117 volts while the other customer is a bit closer and is getting 121 volts. The customer with the lower voltage will average a higher current level for the same amount of power. Is it reasonable to charge that customer more money for the same amount of energy when the difference is due to factors entirely under the control of the power company and over which the customer has no control?

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They don't.
They bill in Watt.hours (or 1000's of Wh = kWh).

They may also bill in V.A.h which are the same as Wh when power factor =1.
As power factor drops (current & voltage increasingly are out of phase) the user gets increasingly-less Watts for the same VA.
BUT the current causes losses in transmission lines and transformers.
At a power factor of zero the customer would consume NO watt.hours but the supplier would still have to send current to them and incur real energy losses in lines and transformers.

Wh = energy.
W = power = rate of using energy.
A = current flow = proportional to W ONLY at fixed V and fixed power factor,
which never occur in real world situatins. So Wh are the real measure of the generation effort and A.h a measure of the line losses.

Wh = energy.

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As for the "less-techy" comparison as stipulated in one of earlier replies, the comparison of electrical flow to water flow can help, handwaving away the finer nuances.

The current (Amps) is the amount of water that would flow, the Voltage is from how high it falls or flows, the Wattage (per hour) is the energy consumed by the user is how fast your useful water-wheel is turning to grind the grain or power some other machine. (And the resistance, Ohms, might be the loss of original flow to obstacles until it gets to your wheel - freefall in vacuum vs. scraping the riverbed, or hitting a dam altogether).

You can have a bucket of water (X Amps) lying still (0 Volts) and so turning nothing (0 Watts => 0 kWh). Or you can pour it from different heights (1 or 100 Volts as in an earlier example) and give more or less spin to the water-wheel over the same amount of time, allowing you to grind more or less grain into flour. As an end user, you are not billed for the bucket of water per se, but for how much grain it helps you to grind in the overall contraption.

In real life, same amount of water falling from different height converts a different amount of its potential energy to kinetical (in water-wheels of the old, or hydro-dam turbines of the modern age - hence the tall dams making large reservoires).

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