# Amps and volts to watts: conversion

When you know the amperage and voltage of an appliance, how do you find the wattage? A heater we were using this evening tripped a breaker on the power strip it was plugged into, and it was a while before I could restore other electrical equipment to regular operation. The heater’s amperage and voltage were given, but not wattage.

Amps times volts equals watts.

Thanx

only for simple (i.e. resistive) loads. Reactive loads like motors, power supplies, and pretty much everything else requires you to know the power factor. Otherwise you don’t know watts, just VA.

If:

1. You know the RMS current (I in amps)
2. You know the RMS voltage (V in volts)
3. You know the power factor of the appliance is 1 (i.e. a purely resistive appliance)

then P = IV where P is the real power in watts (W).

If you do not know the power factor, then P = IV where P is the apparent power in VA.

If the appliance is a heater, then you can assume the power factor is very close to 1.

Well there’s your problem: never plug a high-current device into a cheap power strip, regardless of what its rating is or what standards it supposedly meets (e.g. UL 1363). Those things are junk. At my workplace (an Air Force base), we are not allowed to plug high-current loads into a power strip.

The business about power factors isn’t relevant to your heater, which is a simple resistive device, and so the power factor is 1. If that’s all you’re interested in, stop reading here. However, just for the sake of completeness:

Instantaneous power times instantaneous current is indeed equal to instantaneous power, and a simple average of instantaneous power over time gives the average power. However, the catch is that the instantaneous voltage and current change very rapidly (at 60 Hz, for American electrical systems), and so what’s actually given is a sort of average that’s called the RMS power, for root-mean-square. For a purely resistive device, the instantaneous current is proportional to the instantaneous voltage, which makes the calculation simple: The RMS voltage times the RMS current is equal to the average power.

However, for a capacitative or inductive load, while the current’s amplitude (and hence RMS) value is proportional to the amplitude (and hence RMS) of the voltage, they’re out of synch, so the instantaneous current doesn’t peak at the same time as the instantaneous voltage. This means that the instantaneous power is generally less than it would be for a resistive load, which is where the power factor comes in.

[quote=“Crafter_Man, post:5, topic:740588”]

If:

1. You know the RMS current (I in amps)
2. You know the RMS voltage (V in volts)
3. You know the power factor of the appliance is 1 (i.e. a purely resistive appliance)

then P = IV where P is the real power in watts (W).

If you do not know the power factor, then P = IV where P is the apparent power in VA.

If the appliance is a heater, then you can assume the power factor is very close to 1.

Well there’s your problem: never plug a high-current device into a cheap power strip, regardless of what its rating is or what standards it supposedly meets (e.g. UL 1363). Those things are junk. At my workplace (an Air Force base), we are not allowed to plug high-current loads into a power strip.

[Shrug] cheap it may be, but it has served for nearly ten years powering our TV, DVD player, a cordless phone (high-volume phone supplied by the State of California), a table lamp, and a little digital electric clock. The simple calculation of the power strip’s capacity I estimated at 1860 watts; the electric heater uses 1500.

[Shrug] cheap it may be, but it has served for nearly ten years powering our TV, DVD player, a cordless phone (high-volume phone supplied by the State of California), a table lamp, and a little digital electric clock. The simple calculation of the power strip’s capacity I estimated at 1860 watts; the electric heater uses 1500.
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I would never plug a heater into any power strip. A heater should be plugged into the wall outlet other wise you may get to talk to a fire captain after the fire is put out. Just my 2 cents worth.

I would never plug a heater into any power strip. A heater should be plugged into the wall outlet other wise you may get to talk to a fire captain after the fire is put out. Just my 2 cents worth.
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I read you loud and clear, Snnipe.

I read you loud and clear, Snnipe.
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Good.

For anyone interested in why this is true:

Electrical charge is measured in coulombs. an electron carries an electrical charge of 1.06x10[sup]-19[/sup] coulombs.

Current (the flow of electrons in a conductor) is measured in amperes (amps). One amp equals 6.24x10[sup]18[/sup] electrons flowing past every second. In other words, one coulomb of electrical charge flowing past every second.

Voltage describes the amount of energy carried by a unit of electrical charge. Voltage is measured in units of joules per coulomb. So if an electrical current experiences a drop of 1 volt when it flows through an electrical component, then every coulomb of charge that passes by delivers 1 joule of energy.

Power is energy delivered per unit of time. For electrical devices, the usual unit of power is watts, which is joules per second.

So if you want to calculate power:

P = I * V

Using the units for each of those things, you can see why the units cancel out and leave you with the above tidy equation:

joules/second = (coulombs/second) * (joules per coulomb)

joules/second = (coulombs/electron * electrons/second) * (joules per coulomb)

Nitpick: 1.6x10[sup]-19[/sup].

:smack: Goddamit, I saw that typo as I was writing things up, and I forgot to correct it. Thanks for noting that.

ETA: @Machine Elf’s earlier post.

All of which is true for pure DC and for purely resistive AC loads only.

The formulas still apply 100% to the much more common household situation of non-resistive AC loads, but the complexity is that the I’s & V’s are both modulating out of phase at 60Hz. Plus harmonics. As **Chronos **said farther above.

So as a result an accurate calculation involving a typical AC load requires the use of calculus to blend all the time-varying info together into a single coherent number.

The Power Factor mentioned upthread is a shortcut that for the most common cases gives the calculus result without needing to do the calculus process.

Postscript : I had it pointed out to me (tag affixed to power cord) that the heater should be plugged directly into a wall outlet, never into an extension cord of any kind; fortunately we have one such outlet that isn’t behind an immovable piece of furniture.

I wouldn’t have a problem plugging a heater into an extension cord - provided it was properly rated. Heaters running on 120 VAC are usually around 1500 watts, so they draw about 12 amps. You don’t want to use a shitty little 18-gauge lampcord extension for that; you’d want at least 14-gauge. Most people won’t bother to put that much thought into it though, so the easy advice from heater manufacturers is just to not use an extension cord at all.

Or rather, with someone else having done the calculus for you. I’ve always said that the easiest way to solve a math problem is to realize that somebody else has already solved it.

That said, the calculus is actually remarkably easy, if you’re comfortable with complex numbers.

And I just noticed that there was a mistake in my post #6: “Instantaneous power times instantaneous current is indeed equal to instantaneous power” should say "Instantaneous voltage times instantaneous current is indeed equal to instantaneous power

All 100% true as far as it goes. And so I shall quibble yet again. (Or is that muddy further by trying to clarify?)

For simple non-resistive loads the manufacturer can do the calc once and reduce the calculus to a PF for the end users. For more intricate time-varying loads that’s not 100% true.

As a practical matter for an engineer or electrician sizing a power supply, the PF is plenty close enough for any real world load. As you say.

For somebody trying to understand the physics behind all this, which is what **Machine Elf **seemed to be inviting in post #11, it’s worth noting that even PF is not a complete correction for what’s happening moment to moment with most real world loads. There’s inrush, startup, varying consumption under varying load, etc.
In an AC world, P=IV is a spherical cow; an oversimplified model that obscures as much as it illuminates. P=IV*PF is a darn good approximation, but an approximation nonetheless.

1500 Watts 12.5 amps. I never use 14 gage cords for heavy loads. a 12 gage extension cord is rated at 15 amps. 80% of 15 is 12 so there is no safety margin and I would not do it. I have been down this road many times with tenants in high rise buildings. My final answer to the tenant is that I am willing to have management arrange with the fire marshal to explain the problem. Most of the time the heater gets unplugged and removed. One time management had to explain the facts of life to the tenant.

14-gauge cords in general appear to be rated for 15 amps. To say that there is no safety margin in that current rating ignores the safety margin applied by the manufacturer, and by the NEC. According to the latter, 14-gauge wire is good for 15 amps even with the lowest temperature rating of insulation. A 15-amp rating does not mean that it’s on the verge of melting down at 15 amps; a 12.5-amp load on a cord rated for 15-amp service should be fine.