Convert watts to amps

Paul Mitchell_2

Just a quick question. What is the formula for figuring out amps if all you know is voltage and watts??
Thanks in advance.

Robert O'Connor_3

watts / volts = amps

Alan R. Mercurio_3

Here you go, Paul.

Watts (devided by) Volts = Amps

1800 Watts / 120 Volts = 15 Amps

Your friend in the industry,
Alan R. Mercurio

Oil Tech Talk

Boilerpro_3

That only applies to pure resistance circuits

If I remember my electrical eng. classes. For motors, thngs like Power factor etc, come into play.

Boilerpro

Alan R. Mercurio_3

9:31 - 2:00 Minuets = Joe types faster than Alan


Your friend in the industry,
Alan R. Mercurio

Oil Tech Talk

R. Kalia

Yep, all other replies are incorrect given that the original post did not specify purely resistive loads (or DC power).

You can have lots of AC amps going through a load, with lots of AC volts across that load, but the power can be zero watts if the load is purely inductive or capacititive. Most loads are somewhere in between purely resistive and purely capacititive/inductive, so you need to know the 'power factor' for it.

But the answer obtained when dividing watts by volts is the lower limit for amps; the actual amps can be higher, never lower.

R. Kalia

JimGPE_5

Y'all need to think 8th dimensionally....

SINGLE PHASE RESISTIVE LOADS:

A = Watts/Volts

THREE PHASE RESISTIVE LOADS:

A = Watts/Volts/1.73.

S Ebels

OK, I'll be the dummy here

For those of us who are electrically challenged, please define or describe an example of the following loads and why they draw different amounts of current.

Resistive

Inductive

Capacitive.

Many thanks and a UPS delivery of my brothers best jerky, your choice of flavor, Honey BBQ, Death Valley, Hot Shot, Maple Cherry, Teriyaki or regular, if you can make me understand what you're talking about.

I'll select a winner based on who can make it clear to me and send you 2 lbs if you e-mail me your address.

R. Kalia

can't be done

That's one of those prizes that will remain unclaimed, because it is impossible to explain this coherently with text. You really need graphs and equations and some background training (calculus).

In a resistive load, the current (I) is proportional to the voltage (V): V=IR or I = V/R (Ohm's Law). So as the AC voltage goes up and down, the current wiggles up and down in time with it.

But Ohm's Law is not a real law, because only things that obey Ohm's law (resistors) have to obey Ohm's law. Capacitors and inductors don't have to obey Ohm's Law.

In a capacitor, the current depends not on the voltage but on how fast the voltage is changing with time:
I = C(dV/dt). If the voltage is AC (i.e. a sine function), the current is also AC, but it is not dancing in time with the voltage, but rather is shifted by a quarter cycle. (This follows from calculus, i.e. from differentiating a sine function.) The net result is that sometimes the current is in the same direction as the voltage, sometimes in the opposite direction, so sometimes there are positive watts, sometimes negative watts. It turns out that on the average (averaging over at least one cycle, i.e. at least 1/60th of a second) there are zero watts.

A similar although not identical explanation applies to inductance.

Matt Undy

power factor

First, this is a question that requires a couple chapeters of a circuits or professional wiring book to answer. I would recommned you look at the residential motors and electrical theory chapters of "Practical Electrical Wiring". With that out of the way, I'll try to hit some of the highlights.

Resitive load: Incandescent lamp, iron, heting elemnts in stoves, dryers, and space heaters. Incandescent lamps are a little of a special case because they have high but very brief inrush currents but do not need full voltage at this point as a motor would, so special considerations are only needed in switches, but these are automatically considered in modern UL switch standards.

Power Factor:

In motor loads and some other types of loads, the laod acts as a capacitor and an inductor. This means that they can store up some energy andrelease it at some other part of the voltage cycle. AC power consists of a voltage that fluctuates between one polarity and its opposite, its a constant rate of change, so a plot of voltage over time is a sine wave. When a device store up energy and releases it as the voltage cycles, it may(and udually doesn't) draw its maximum current at the time of maximum voltage so the product of this current and voltage waveform can be less than if the paks were aligned. Wire still heats based on this maximum current, so the wire must be sized to handle it, not necisarily just the apearent power. This usually isn't a problem in residential wiring where there are few motor loads, the code requires that permenently wired device such as motors only load the circuit to 80% of its rating(for a dedicated circuit. read the code, don't assume anything about what is and isn't suitable from my explinations), and most residential motors are relitively small. Power factor is a major problem in industrial installations.(I think my description of why power factor occurs is correct, please let me know if I am wrong.)

motor loads: induction motors(capacitor start, split phase, 3 phase) draw current in proportion to how close their rotor speed is to its ideal speed. This means that the curent they draw when their rotor hasn't started spining yet and while it is coming up to speed is much grater than when it is runing. This also means that the wiring needs to be sized such that the voltage drop isn't too great at this higher current (sometimes as much as 10 times the current when the motor is dleivering its rated horsepower at rated speed). Since this higher current delivers the power to start the load connected to the motor's shaft, the voltage needs to be kept near the motor's rated voltage so that sufficient power is still delivered to the motor at the high startign current. This high starting current is why switches and relays are rated differently for motor loads and resitive loads.

Matt

scrook_2

for power factor not = 1

W/V*(PF)=Amps

where W is power in Watts, V is voltage in volts and PF is power factor (0 to 1 where 1=100%PF and 0=0%PF)

A resistive load will be 100% PF or 1, a motor, fluoresent lamp ballast, etc. will be somewhat less, maybe 0.9, maybe 0.6, etc. If you know the PF you are in business, if you don't life is rough, as it can be a pain to measure w/o a special meter that measures both the AC current and the AC voltage and the relationship between the two.

If you have an ideal inductor (or coil) the current and voltage flow out of phase (The peak current and the peak voltage in the AC cycle come at opposite times) and no work is done so no power (in the rigid physics definition of the word) is used/extracted, and the PF = 0

Likewise (but less common) if you have an ideal capacitor the same thing happens. A motor looks part like like a coil and part like a resistor, so some of the current contributes to power (as electricity in and as rotation out) and some doesn't. The power company hates that current that does no work yielding no power because the watt-hour meter doesn't measure it (which is the way it should be because it delivers no power to you) BUT the current does heat up their transformers and wires (because they are not perfect) so the electric co has to install bigger transformers and wires then they might have to, or must install capacitor to offset the inductive effect of the motors etc. which costs them money.

Power factor measures how much of the current is contributing to power and how much is "reactive power" the useless part

V*A=VA (Volt-Amps)
V*A*PF=W (Watts)
For all but pure resistive loads Volt-Amps will be bigger than Watts

Paul Mitchell_2

I'll give it a shot in the dark

Resistive draws more current because of the resistance it causes requiring more current to complete the circuit or provide voltage.

Inductive draws less current because of less resistance.

Capacitive has little current but higher voltage because of storage capacity.

Just lookin for some jerky....no clue what I am talking about..

Glen

I asked Timmie -

watts equal volt amps. 5 w = 5 va. It cured me from looking further. Thanks Timmie.

bob_25

I can,t believe

that no one mentioned "ELI the ICE man" . Where E is volts, I is current, L is inductance and C is capacitance. In a circuit with inductance the voltage peaks before the current ELI and with capacitance the current peaks before the voltage ICE. There is resistance you can measure with an ohm meter and impedance caused by inductive reactance which is an aparent resistance caused by a loop in a wire. Ever wonder why the motor in a Honeywell zone valve doesn't burn out when it sits there stalled in the open position. It's because the motor is wound with so much inductive reactance it can't draw enough current to hurt it's self. You can kind of think of inductive reactance (apparent resistance) caused by a loop in a wire the same as the increased resistance to flow you get when you put a bend in a pipe. Capacitive reactance is the opposite, it's like a bladder expansion tank it trys to maintain pressure(voltage) for an instant when system pressure drops.

joe_17

ELI the ICE man...

Ah remember him well.I was waiting for somebody to mention him.Volts lead Current in an Inductive load.Current leads Volts in a capacitance load.Dont have to worry about resistance load because volt and current are in phase.

Harold

Graphic explanation

Along the line of remembering real far back.

Get yourself a piece of paper and a piece of transparant plastic to go over it.

Draw a sine wave on each one. Draw on the paper and then trace it on the plastic.

One sine wave will represent voltage and the other current. They can be different sizes but make them the same for the exercise. When they are directly over each other they represent a resistive load. If you assign a peak value to the two sine waves of say 120 V and 10 A, the product of the two at any point on the waveform is the power in watts. At 50% down the curve the power is 50% of full value (P=EI).

Now slide the plastic one way or the other. This represents a phase angle change caused by inductive or capacitive load. Take a ruler and put it perpendicular to the sinewaves. The two points where the ruler crosses the sinewaves is what you multiply. The same method of calculating power applies but now the result is changed from the resistive value by the amount of shift of the phase.

You can see that if you stay with the same POWER (watts) and keep the line voltage the same then your current waveform has to get bigger to provide the same product when the curve values are multiplied together. For the specific phase diference (power factor is the normalized phase diference) between the waveforms (characteristic of the specific device) this is the Volt Amp product that you see in some equipment ratings. This is a prefered rating for things like transformers and generators becaues it deals with non-resitve load sizing.

gehring

So for the purpose of sizing single phase residential AC circuits is it safe to say: WATTS/VOLTS = AMPS/.75 = safe circuit capacity??

In other words if I calculate my connected electrical load on a circuit to be 1800 Watts at 120 Volts, then 1800/120 = 15 Amps divided by .75 = 20 Amps. I would run No. 12 wire to a 20 Amp breaker for that circuit.

jerry scharf

going in circles

Paul,

If you think everyone has you going in circles, you're right. When you start with something going in circles and look at what shows up on a line, you get that wiggling line called a sine wave. So some AC generator is turning in circles, and what happens is that the electrons move back and forth in a wire, thus all the talk about sine waves. For me, it's always easier to see the circles.

So there are three things going on that all of electricity deal with. Voltage is how hard you are pushing on the electrons. Current is how fast the electrons are going. Impedence is how hard it is to push the electrons.

If you imagine going around the circle and you're pushing on the electrons. Then the electrons flow directly related to the push, and that's resistance, Ohm's law and all that. If you put a pair of plates near each other, they store electric charge. That's capacitance, and it's the rate of change of the voltage that causes it to happen. This means it's 90 degrees away from you acting as the pusher. If you build a coil, now you're storing magnetic energy, that's inductance. This is 90 degrees away from you the other direction from capacitance.

The ratio of the power consumed (watts) to the voltage*current product (VA) is the power factor. If you can balance out the capacitance and inductance, then you have the minimum current to produce a given wattage.

On power grids, more of the load is resistive or inductive, so the power company puts big capacitors at it's substations to minimize the current they use.They also step the voltage way up, so they can reduce the current and thus the line loss.

Or you could take the other approach I have seen people use to attempt mastry of this. Buy an appropriate text book, put it under your pillow and go to sleep. Come to think of it, that sounds a bit like capacitance.

still going in circles, jerry

Mike T., Swampeast MO

Loads & Power

Here's my admittedly simple take:

Resistive: Example--electric baseboard heater. Power drawn is simply a function of the resistance (ohms) of the heating element. Alternating or direct current doesn't matter as at a given voltage, it will draw the same amount of amps (current).

Inductive: Example--Kirby Vacuum Cleaner. This type of device works by magnetic induction (the inducing of current between devices NOT connected). Works ONLY with alternating current. The unconnected devices are the rotor (moving) and stator (stationary). The alternating current enters one of these devices and creates a magnetic field that expands and collapses with the frequency of the line current. As this happens, a magnetic field is induced in the other device--however the coils in both devices are designed in a sort of spiral that induces a ONE-WAY "twist" in the magnetic field and the rotor turns--in one direction only. Without the "twist" of the expanding/collapsing field, the motor would just oscillate back-and-forth with the frequency of the current. You can see the "twist" by looking at the way the coils are wound.

Understanding how much current is drawn through such a device is difficult as it "depends". It depends both on the speed (remember--this simple motor does not have a built-in way to maintain a set speed so speed varies with load) and the amount of usable work produced. This may be the wrong way to think of it, but if you consider "speed" to be "voltage" and "work" to be "amperage" it makes sense to me...

On that Kirby vacuum: The "normal" power draw is with the unit both moving air and moving the rotating brush. The effort required to spin the brush takes a lot of work (amperage)--BUT it takes so much work (amps) to turn the brush that speed (voltage) decreases. As the speed (voltage) decreases, so does the ability of the device to produce work and there is a corresponding drop in amperage. (Don't have one here but the "normal" amp draw of a Kirby vacuum is say 3.5 amps.) This "normal" rating is an attempt to get the most work for the least amount of amperage...and is represented as the "power factor"--in this case quite high.

Remove the rotating brush and you've removed a lot of work--the speed (voltage) increases greatly. As the speed (voltage) increases the work (amperage required to move air) increases as well. The vacuum will now have both high speed (voltage) and high work (amperage) and will draw say 5 amps. BUT, since the speed (voltage) is so high, the device is not as efficient at producing work and the power factor of the motor is changed.

Now cover the hose with your hand. Work (amperage) plummets and speed (voltage) increases. You hear a high-pitched whine and think that the motor is working extremely hard--even drawing more amps, but it's not. Why? While speed has increased, the only work being done is actually turning the fan, the unit will only be drawing 2 amps. Power factor becomes essentially meaningless because you are accomplishing no useful work.

Now let the brush bearing get filled with pet hair. Power required to turn the brush climbs and speed drops. Work (amperage) increases faster than speed (voltage) decreases. The amps drawn might increase to 4.5, but you are getting less usable work--essentially "eating" your power factor.

Capacitive: Example--an audio amplifier (hope that's a good choice). Capacitors store electrical energy. Once "full" they essentially stop drawing current--in this way they "block" direct current. Power consumed depends on how fast and how much of this stored energy you consume. An audio amplifier essentially stores electric energy in its big capacitors and "lets it out" as determined by the setting of the volume control. The power used is very low at idle and as high as the ability to replenish the energy supply in the capacitors when you are "head banging".

bob_25

Current

Jerry, current is a flow rate like GPM. It's defined as 62800000000000000000 electrons past a fixed point in a wire in one second. bob

Paul Mitchell_2

Just think

I just wanted to get the amp rating on a 4500 watt electric hwh. I appreciate all the tech talk.
Thanks

S Ebels

That's what I was looking for

E-mail me your address Mike.

Harold

easy

For a 4500 W water heater the load is pure (for all intents and purposes) resistive so P=E*I. Or I=P/E If the heater is 240V it draws 18.75 Amps.

Mike T., Swampeast MO

Will still say that the overwhelming majority of people (including me) don't fully understand how A/C motors actually work.

Dear Mr. Tesla's designs were and still are perfection embodied--the only "improvements" have come from enhanced materials and manufacturing techniques.

ALL of his original motor designs remain essentially unchanged and he developed motors for all purposes:

1) Shaded pole (the type with stacks of plates and a thick wire running at a strange angle across): Easy to make and inexpensive, but not very efficient.

2) Single phase induction: More expensive to manufacture, higher power-to-weight ratio than shaded pole and able to be powered with a single source of A/C current.

3) Polyphase induction: Extremely efficient (large ones with 3 or more phases approach a 1-1 power in/power out ratio) but they do require more wires...

4) Reversible induction: Extremely complex and expensive to manufacture, but still less expensive than two motors designed to operate in opposite rotation.

5) Synchronous: They either turn at a given speed or they don't turn. Again, complex and expensive (I believe they use multiple coils that are powered in a "chase" fashion), but required for things like timers. "Stepper" motors used in computers are (I believe) just synchronous motors with digitally controlled "chasing" of the coils.

-------------------------------------

Look up Tesla but don't get too involved in the "mad scientist" aspect. He could "see" alternating current in a way that few (if any) have ever seen it--similar perhaps to the way I seem to "see" simple radiation and convection. Now if only I can learn to "see" polyphase radiation, I will understand how to collect the diffuse energy of the cosmos, concentrate it and use it without wires. But if I or anyone ever actually understand this, industry will make it just as impossible today as it was for Tesla...

Mike T., Swampeast MO

The jerky

Sent my address but forgot to include jerky variety.

"Death Valley" sounds interesting, but I'm not sure what it means.

I do like pepper hot, but not mustard or curry hot...

Straight jalape

JimGPE_6

Okay

You got me laughing out loud with that one!

JimGPE_6

Yeah,

You pretty much have to tell us that up front or we'll tell you how that watch is made, why its made that way, how long they've been made that way, who's idea it was to make them that way, no, really it was this OTHER guy's idea, etc., etc.

We're like that around here.

Matt Undy

> Here's my admittedly simple take:

>

> Resistive:

> Example--electric baseboard heater. Power drawn

> is simply a function of the resistance (ohms) of

> the heating element. Alternating or direct

> current doesn't matter as at a given voltage, it

> will draw the same amount of amps (current).

> Inductive: Example--Kirby Vacuum Cleaner. This

> type of device works by magnetic induction (the

> inducing of current between devices NOT

> connected). Works ONLY with alternating current.

> The unconnected devices are the rotor (moving)

> and stator (stationary). The alternating current

> enters one of these devices and creates a

> magnetic field that expands and collapses with

> the frequency of the line current. As this

> happens, a magnetic field is induced in the other

> device--however the coils in both devices are

> designed in a sort of spiral that induces a

> ONE-WAY "twist" in the magnetic field and the

> rotor turns--in one direction only. Without the

> "twist" of the expanding/collapsing field, the

> motor would just oscillate back-and-forth with

> the frequency of the current. You can see the

> "twist" by looking at the way the coils are

> wound.

>

> Understanding how much current is drawn

> through such a device is difficult as it

> "depends". It depends both on the speed

> (remember--this simple motor does not have a

> built-in way to maintain a set speed so speed

> varies with load) and the amount of usable work

> produced. This may be the wrong way to think of

> it, but if you consider "speed" to be "voltage"

> and "work" to be "amperage" it makes sense to

> me...

>

> On that Kirby vacuum: The "normal"

> power draw is with the unit both moving air and

> moving the rotating brush. The effort required

> to spin the brush takes a lot of work

> (amperage)--BUT it takes so much work (amps) to

> turn the brush that speed (voltage) decreases.

> As the speed (voltage) decreases, so does the

> ability of the device to produce work and there

> is a corresponding drop in amperage. (Don't have

> one here but the "normal" amp draw of a Kirby

> vacuum is say 3.5 amps.) This "normal" rating is

> an attempt to get the most work for the least

> amount of amperage...and is represented as the

> "power factor"--in this case quite

> high.

>

> Remove the rotating brush and you've

> removed a lot of work--the speed (voltage)

> increases greatly. As the speed (voltage)

> increases the work (amperage required to move

> air) increases as well. The vacuum will now have

> both high speed (voltage) and high work

> (amperage) and will draw say 5 amps. BUT, since

> the speed (voltage) is so high, the device is not

> as efficient at producing work and the power

> factor of the motor is changed.

>

> Now cover the

> hose with your hand. Work (amperage) plummets

> and speed (voltage) increases. You hear a

> high-pitched whine and think that the motor is

> working extremely hard--even drawing more amps,

> but it's not. Why? While speed has increased,

> the only work being done is actually turning the

> fan, the unit will only be drawing 2 amps. Power

> factor becomes essentially meaningless because

> you are accomplishing no useful work.

>

> Now let

> the brush bearing get filled with pet hair.

> Power required to turn the brush climbs and speed

> drops. Work (amperage) increases faster than

> speed (voltage) decreases. The amps drawn might

> increase to 4.5, but you are getting less usable

> work--essentially "eating" your power

> factor.

>

> Capacitive: Example--an audio

> amplifier (hope that's a good choice).

> Capacitors store electrical energy. Once "full"

> they essentially stop drawing current--in this

> way they "block" direct current. Power consumed

> depends on how fast and how much of this stored

> energy you consume. An audio amplifier

> essentially stores electric energy in its big

> capacitors and "lets it out" as determined by the

> setting of the volume control. The power used is

> very low at idle and as high as the ability to

> replenish the energy supply in the capacitors

> when you are "head banging".

Matt Undy

> Will still say that the overwhelming majority of

> people (including me) don't fully understand how

> A/C motors actually work.

>

> Dear Mr. Tesla's

> designs were and still are perfection

> embodied--the only "improvements" have come from

> enhanced materials and manufacturing

> techniques.

>

> ALL of his original motor designs

> remain essentially unchanged and he developed

> motors for all purposes:

>

> 1) Shaded pole (the

> type with stacks of plates and a thick wire

> running at a strange angle across): Easy to make

> and inexpensive, but not very efficient.

>


These are only used in applications where a very small amount of power is needed such as a range hood or bathroom exhaust fan or a draft inducer blower.

> 2)

> Single phase induction: More expensive to

> manufacture, higher power-to-weight ratio than

> shaded pole and able to be powered with a single

> source of A/C current.


Theese are trivially reversable if the start and run windings are both consturcted to be able to carry the full load current of the motor. They may be reversed by simply reversing the start wainding. The run winding produces a push with evey cycle of the AC line, but this push will only vibrate the armeture unless it is already turning. The start winding provides the out of phase push (either through using a capacitor to shift the phase of the AC or through positioning of the start and run windings.) to start the motro turnign and to determine which direction it turns in. If you ever encounter a motor that just hums but doesn't turn until you give it a bit of a turn, it has a problem with the start circuit. A centrufigual switch turns off the start winding as the motor aproaches speed. The start capacitors commonly fail, if it has a capcitor and it exhibits the above symptoms it is probably the capacitor.

Universal or AC/DC motors have an even higher power to weight ratio and are about the only ones suitable for hand held power tools.

>

> 3) Polyphase

> induction: Extremely efficient (large ones with

> 3 or more phases approach a 1-1 power in/power

> out ratio) but they do require more

> wires...

>

> 4) Reversible induction: Extremely

> complex and expensive to manufacture, but still

> less expensive than two motors designed to

> operate in opposite rotation.


I'm not sure what this means. A 3 phase induction is reversed by reversing 2 of the pahses and a single pahse induction is reversed by swaping the start and run windings(or some other swithc of the start winding. I'm nto positive its swaping the start and run but virtually any single phase induction(except OEM motors) is reveresed by swaping 2 wires.)
>

> 5) Synchronous:

> They either turn at a given speed or they don't

> turn. Again, complex and expensive (I believe

> they use multiple coils that are powered in a

> "chase" fashion), but required for things like

> timers. "Stepper" motors used in computers are

> (I believe) just synchronous motors with

> digitally controlled "chasing" of the

> coils.

>

> -------------------------------------

> Look up Tesla but don't get too involved in the

> "mad scientist" aspect. He could "see"

> alternating current in a way that few (if any)

> have ever seen it--similar perhaps to the way I


He understood calculus, most people at the time, most notably Edison didn't

> seem to "see" simple radiation and convection.

> Now if only I can learn to "see" polyphase

> radiation, I will understand how to collect the

> diffuse energy of the cosmos, concentrate it and

> use it without wires. But if I or anyone ever

> actually understand this, industry will make it

> just as impossible today as it was for Tesla...

JimGPE_5

Only if its D/C ...

The really bizzarre part is if its A/C, the same 62,800,000,000,000,000,000 electrons go past the same point over and over again, back and forth, back and forth. You don't actually GET anything from the electric company, they just jiggle the electrons you bought when you bought your wiring!

Mike T., Swampeast MO

Will get even crazier...

Those 62,800,000,000,000,000,000 electrons came from the sun and you're just "jiggling" solar energy that had to make its way from the sun via the atmosphere via lightning...

Wouldn't it be more fun (and efficient) to "jiggle" those electrons between the atmosphere and the earth by ourselves?