# Amps vs. Watts vs. Volts?



## cmeisenzahl (Jan 4, 2003)

Can someone point me to an "electronics for dummies" type page, or explain this to me? I still after all these years don't understand the difference between these. My dad has tried to explain it to me several times with little success. :-(

Thanks in advance!


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## Nerd (Jan 4, 2003)

Lemme have the honour of taking up the challenge. Let's take electricity as water. Wire as garden hose.

Voltage is amount of pressure. So more voltage, more pressure, water shoot out of hose faster.

Ampers is amount of water. Bigger hose, more water can flow at one time.

To make it as unconfusing as possible, take your normal everyday garden hose: You pinch it, you get higher voltage, you change hose to bigger hose, more water can flow, but that doesn't = higher voltage because water may flow slower instead.

Watts = total amount of water flowing and at what speed = Volts x Amps.

Tell me which part do you not understand and I'll see if I can simplify it some more.


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## luxO (Jan 4, 2003)

"To make it as unconfusing as possible, take your normal everyday garden hose: You pinch it, you get higher voltage, you change hose to bigger hose, more water can flow, but that doesn't = higher voltage because water may flow slower instead."

No, pinching it adds resistance, voltage stays the same, current drops.


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## Empath (Jan 4, 2003)

The amount of worked performed by a circuit today was 100 watts.

It did it with 100 amps. It didn't take much force since it had so many amps. It only took 1 volt.

If there were only 50 amps, it would take twice as much force. It would take 2 volts.

If there were only 25 amps, it would take four volts.

............................

My men lifted 100 lbs today (watts). Their labor contract said they were only allowed to lift 1 pound each (amps). It took 100 men (1 volt each) to lift the 100 pounds.

The labor union renegotiated, and they can now lift 2 pounds each (amps). It now only takes 50 men (2 volts each) to lift the 100 pounds.

I'm trying to renegotiate for them to lift 4 pounds each (amps). That way, it will only take 25 men (4 volts each) to lift the 100 pounds.

........................

Watts (the work done) = Amps (the resources consumed) times Volts (the strength of the resource units)


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## McGizmo (Jan 4, 2003)

Let me throw this analogy at you:

Electrons do the work and the more that come through a gate and the speed at which they pass through, the more work (watts) can be done. The gate restricts how many electrons can pass through at the same time. The gate (wire) size dictates this number and think of this number as amps. Now the electrons can flow through the gate at different speeds (voltage) and the faster they go through, the more energy or power they impart (Watts).

Watts = Amps X Volts

It helps me to think of the electrons moving through the wire as I can see that amperage capacity is then a function of wire size (how many electrons can fit in the gate or tunnel cross section) and then I visualize the speed at which they are moving as the voltage or velocity.

To use this analogy further, consider a resistor or resistance as speed bumps




The electrons are slowed down at this point and a voltage drop occurrs. As the electrons hit the speed bumps and are slowed, energy is lost.

Another reason I like to think of voltage being the velocity of the electrons is that I can visualize electrons speeding through a wire and if they come to a gap in the wire, the faster they are moving, the farther I can see them successfully jumping across this gap (arc). The greater the voltage, the greater the arc.

The annoying static electricity is a result of a few, misbehaved electrons, going well in excess of the speed limit and doing Evil Kenevil stunts at our expense.


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## cmeisenzahl (Jan 4, 2003)

Thanks very much guys!!


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## Empath (Jan 4, 2003)

Chris,
Try to think in literal terms.

Watts: The amount of work to do.

Amperage: The resources (the number of electrons)you're working with.

Voltage: The power (the muscle of each electron)of each unit of your resources.


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## JackBlades (Jan 4, 2003)

Ask 'em about resistor values now.


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## logicnerd411 (Jan 4, 2003)

Ooh, the pretty colors...


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## Jbirk (Jan 5, 2003)

Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Gray
White


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## Monsters_Inc (Jan 5, 2003)

10kohm
1kohm
100ohm
10ohm
1ohm
1milliohm...

am i anywhere near the right track?


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## luxO (Jan 5, 2003)

"am i anywhere near the right track?"

um, no.


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## INRETECH (Jan 5, 2003)

What is Electricity? 
By Dave Barry 

What in the world is electricity? 

And where does it go after it leaves the toaster? 

Here is a simple experiment that will teach you an important electrical lesson: On a cool, dry day, scuff your feet along a carpet, then reach your hand into a friend's mouth and touch one of his dental fillings. Did you notice how your friend twitched violently and cried out in pain? This teaches us that electricity can be a very powerful force, but we must never use it to hurt others unless we need to learn an important electrical lesson. 

It also teaches us how an electrical circuit works. When you scuffed your feet, you picked up batches of "electrons," which are very small objects that carpet manufacturers weave into carpets so they will attract dirt. The electrons travel through your bloodstream and collect in your finger, where they form a spark that leaps to your friend's filling, then travels down to his feet and back into the carpet, thus completing the circuit. 

Amazing Electronic Fact: If you scuffed your feet long enough without touching anything, you would build up so many electrons that your finger would explode! but this is nothing to worry about unless you have carpeting. 

Although we modern persons tend to take our electric lights, radios, mixers, etc. for granted, hundreds of years ago people did not have any of these things, which is just as well because there was no place to plug them in. Then along came the first Electrical Pioneer, Benjamin Franklin, who flew a kite in a lightning storm and received a serious electrical shock. This proved that lightning was powered by the same force as carpets, but it also damaged Franklin's brain so severely that he started speaking only in incomprehensible maxims, such as "A penny saved is a penny earned." Eventually he had to be given a job running the post office. 

After Franklin came a herd of Electrical Pioneers whose names have become part of our electrical terminology: Myron Volt, Mary Louise Amp, James Watt, Bob Transformer, etc. These pioneers conducted many important electrical experiments. For example, in 1780 Luigi Galvani discovered (this is the truth) that when he attached two different kinds of metal to the leg of a frog, an electrical current developed and the frog's leg kicked, even though it was no longer attached to the frog, which was dead anyway. Galvani's discovery led to enormous advances in the field of amphibian medicine. Today, skilled veterinary surgeons can take a frog that has been seriously injured or killed, implant pieces of metal in its muscles, and watch it hop back into the pond just like a normal frog, except for the fact that it sinks like a stone. 

But the greatest Electrical Pioneer of them all was Thomas Edison, who was a brilliant inventor despite the fact that he had little formal education and lived in New Jersey. Edison's first major invention in 1877, was the phonograph, which could soon be found in thousands of American homes, where it basically sat until 1923, when the record was invented. But Edison's greatest achievement came in 1879, when he invented the electric company. Edison's design was a brilliant adaptation of the simple electrical circuit: The electric company sends electricity through a wire to a customer, then immediately gets the electricity back through another wire, then (this is the brilliant part) sends it right back to the customer again. 

This means that an electric company can sell a customer the same batch of electricity thousands of times a day and never get caught, since very few customer take the time to examine their electricity closely. In fact the last year any new electricity was generated in the United States was 1937; the electric companies have been merely re-selling it ever since, which is why they have so much free time to apply for rate increases. 

Today, thanks to men like Edison and Franklin, and frogs like Galvani's, we receive almost unlimited benefits from electricity. For example, in the past decade scientists developed the laser, an electronic appliance so powerful that it can vaporize a bulldozer 2,000 yards away, yet so precise that doctors can use it to perform delicate operations to the human eyeball, provided they remember to change the power setting from "Vaporize Bulldozer" to "Delicate."


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## x-ray (Jan 5, 2003)




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## 0tr3b (Jan 6, 2003)

> Originally posted by Empath:
> *The amount of worked performed by a circuit today was 100 watts.
> 
> It did it with 100 amps. It didn't take much force since it had so many amps. It only took 1 volt.
> ...


<font size="2" face="Verdana, Arial">Just to clarify it a little bit, watt is a unit of power. The actual work done is measured in watt-hours or watt-seconds. High power makes quick work.

aa


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## Wingerr (Jan 6, 2003)

That Dave Barry guy is a scream- Calvin and Hobbesque, only more so..


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## Willmore (Jan 8, 2003)

Short form:
V is voltage in, uhh, volts
I is current in Amperes
R is resistance in ohms
W is power in watts

Ohms law:
V=I*R
Power law:
W=V*I or W=I^2*R (using ohms law)

The rest is just different ways of arranging these equations.


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## dharmil007 (Jun 3, 2012)

Thank You People for such a nice & wonderFul explanation.
i am on my way to understand this things.

Still i need a few clarifications, pls. can anyOne help me out.


1} In home sockets/plugs voltage is constant {240V in india}, so we just need to calculate & AmPs are given so we just need to calculate Power {Watts} with
formula P = V*I.
But sometimes this formula doesnt give accurate results.
I just saw my SMPS, which is 450W & has 230V & 5A.

Now according to the formula:
V*I = W
230*5 = 1150W






This doesnt get equal to the one mentioned on the SMPS.

y Is this so ??:thinking:


2} In Mobile phones or any device battery operated, how do we calculate electric consumption.
'coz it is measured in mAh {1000mAh = 1Ah}, but it has variable voltage.
& a lot others other things.

its just a lot of confusing.

Can someone pls explain in simple manner ?


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## HKJ (Jun 3, 2012)

dharmil007 said:


> Can someone pls explain in simple manner ?



I am afraid that there is no simple explanation.

With AC power the formula is not P=V*I, but need an added Power Factor (PF), this can also sometimes be called "cos phi".
I.e. the formula is P=V*I*PF
The PF is between 0 and 1.
For resistors the PF is 1, for most other stuff it is less than 1. Some equipment is marked with the PF (or cos phi).


When looking at ratings on the back of equipment the current is often specified to high. This value is used when you calculate the mains fuse, to avoid it blowing when you turn the equipment on.


Battery supplied equipment does vary in how they draw their power:
1) Some will draw a constant current, independent of voltage
2) Some will increase current draw when voltage drops, i.e. using nearly constant power.
3) Some will reduce current draw when voltage drops.
If you look in my flashlight reviews you will see a chart showing current and power draw, depending on voltage.


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## TedTheLed (Jun 3, 2012)

I discovered something interesting the other day.

Electrons move very slowly through wire!

Its not the electrons doing the work so much as the magnetic field.

The electrons appear to move at or near the speed of light, say, when you flick a switch, and the light comes on immediately, because they are all packed in the wire 'shoulder to shoulder' and pushing the first electron in line causes the last electron in line to move..

but the speed of the electron is measured in feet per second! hard to believe, I know, counter-intuitive too, but check it out, Don.


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## SPNKr (Jun 3, 2012)

dharmil007 said:


> Can someone pls explain in simple manner ?



There are two ways of transmitting electricity through wires, direct and alternating current, DC and AC in short, you may have come across them. Direct current is where the direction of the current (travel of electrons) doesn't change, and alternating current is where the direction alternates. Imagine a straight wire. DC is where the current always flows to the left, for instance, whereas in AC the current will alternate in flow from right to left. 

DC current is used in small electrical devices powered by batteries, so the P=IV equation can be used directly. AC is used by power plants to transmit power to your sockets due to the ability of its voltage to be stepped up and down via transformers, reducing power loss during transmission (don't worry about this, it's not really that relevant to your question). The thing is since the current and voltage in an AC circuit always vary, they cannot be plugged directly into the P=IV equation. Instead, we have to take something called the root mean square value of the current and voltage, calculated by squaring the the values of one cycle of the oscillation, taking the mean (I won't go into detail) and then square-rooting it. Basically, this would give a value of current and voltage that is equivalent to the values in a DC circuit that does the same amount of work (i.e. power).

Now as far as I know the AC current that we get from the mains follows that of a sine curve. Meaning the potential difference varies in a way that plotting the voltage against time on a graph would yield a sine curve. To obtain the RMS values for this type of AC circuit, it is simply dividing the maximum current or voltage by the square root of 2. The reason behind this will become clear when you see the graph. Therefore, to obtain the power of an AC circuit, meaning anything that's connected to the mains, the equation to follow is P=I(RMS)V(RMS), or P=[I(max)V(max)]/2.

Try that and you should get the correct values in your picture.

Regarding your second question, mAh represents how long a battery can sustain a particular current. 1000 mAh just means the battery can sustain a current of 1000 milli-amperes for one hour, 500 milli-amperes for two hours etc. Current alone isn't enough in determining how long your battery will last, you'll also need the voltage, and using that you can determine the amount of energy the battery has stored. Since it is a lithium ion battery, the potential difference between the two terminals is (for convenience's sake), 3.7V. So using P=IV, the battery has stored 3700 milli-watt hours of energy (power is work done per unit time). That means your battery has enough stored energy to keep a machine that draws 3700 milli-watts of power running for one hour. It's very very tricky to try to use this number to predict how long the battery will last, as like you mentioned, the power consumed by the phone varies according the usage and the PD across a Li-ion battery in reality isn't always 3.7V. However, if, say your phone lasts 10 hours on a single charge, you can estimate that it has consumed, on average, 370 milli-watt hours of energy, per hour


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## dharmil007 (Jun 3, 2012)

SPNKr said:


> Now as far as I know the AC current that we get from the mains follows that of a sine curve. Meaning the potential difference varies in a way that plotting the voltage against time on a graph would yield a sine curve. To obtain the RMS values for this type of AC circuit, it is simply dividing the maximum current or voltage by the square root of 2. The reason behind this will become clear when you see the graph. Therefore, to obtain the power of an AC circuit, meaning anything that's connected to the mains, the equation to follow is P=I(RMS)V(RMS), or P=[I(max)V(max)]/2.
> 
> Try that and you should get the correct values in your picture.



i Treid calculating, but it dint gave the answer.
[I(max)*V(max)]/2
[5*230]=1150/2=575W
But the SMPS has 450W ??:thinking::thinking::thinking:




SPNKr said:


> Regarding your second question, mAh represents how long a battery can sustain a particular current. 1000 mAh just means the battery can sustain a current of 1000 milli-amperes for one hour, 500 milli-amperes for two hours etc. Current alone isn't enough in determining how long your battery will last, you'll also need the voltage, and using that you can determine the amount of energy the battery has stored. Since it is a lithium ion battery, the potential difference between the two terminals is (for convenience's sake), 3.7V. So using P=IV, the battery has stored 3700 milli-watt hours of energy (power is work done per unit time). That means your battery has enough stored energy to keep a machine that draws 3700 milli-watts of power running for one hour. It's very very tricky to try to use this number to predict how long the battery will last, as like you mentioned, the power consumed by the phone varies according the usage and the PD across a Li-ion battery in reality isn't always 3.7V. However, if, say your phone lasts 10 hours on a single charge, you can estimate that it has consumed, on average, 370 milli-watt hours of energy, per hour



Thanks for the explanation.
But, This is not what i meant in the question
Maybe i was too confused mySelf, so coudnt put it in proper terms.

What i meant was :
Voltage is used in supply.
Then in mobile phones why it is the term that is used for drainage & it varies. ??






& another question which i came through.

i Have an AC, the trigger used for AC is of 25AmPs.
When i used multimeter, it showed that while turning on AC, Amp had gone to 20AmPs & gradually came down to 10-12AmPs.
So according to the formula my AC uses 2900W {240*12} or 2.9kW.
i Need to calculate this on the input {trigger} or Output ???

& this is how my Electric bill is calculated ???


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## SPNKr (Jun 3, 2012)

I just did a little reading up, and found out 230V actually refers to the RMS value. I'm not sure why we don't get 450W, it could be that 450W refers to the output power, and some power is lost to resistance in the power supply. It could also be that it's mislabeled.

Voltage is the potential difference between 2 points on a circuit, and in your phone it's referring to the PD between the two terminals of the battery. As I've mentioned before the terminal voltage of the battery varies according to how much it's charged, this is due to the chemistry of the battery. Li-ions are 4.2V when fully charged and the voltage slowly decreases as they're discharged. The voltage measurement is a way of telling you how full your battery is.

I'm not quite sure what you mean by trigger, but if you're trying to measure the amount of work your device can do, you use the output. If you're interested in your electricity bills, measure the input. This is because the input of your devices is essentially the output of the power plant to your home, and they'll charge according to that.

I just noticed this thread is 9 years old O.O


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## HKJ (Jun 3, 2012)

As I wrote above: You cannot just multiply voltage and current to get power when working with AC, it only works for resistive loads (except if you do it fast enough)!

This also means that you need special meters to measure power when working with AC, no ordinary DMM can do it.
The meter needs to multiply voltage and current in real time and calculated the average of that. For digital meters that means measuring and multiplying voltage and current a few 1000's times each second.

The green meter below can do that, the two other cannot:


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## Mr Happy (Jun 3, 2012)

HKJ said:


> The green meter below can do that, the two other cannot:



The two yellow meters shown there are quite expensive. The green meter (a Metrahit Energy?) is _very_ expensive! :green:


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## peterharvey73 (Jun 3, 2012)

Out of interest, what is the model number of the Fluke on the left?
Is it still commercially available?


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## HKJ (Jun 3, 2012)

Mr Happy said:


> The two yellow meters shown there are quite expensive. The green meter (a Metrahit Energy?) is _very_ expensive! :green:



Yes, it is the Energy and the ability to measure Watt was the main reason I bought it (It can also measure the PF I wrote about). I do have other DMM's, even some more expensive, but none of them can measure Watt.
There are a few other meters that can do Watt, but they are rather limited.



peterharvey73 said:


> Out of interest, what is the model number of the Fluke on the left? Is it still commercially available?



It is the 189 and the other Fluke is the 289 that has replaced the 189 in the Fluke lineup.


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## peterharvey73 (Jun 3, 2012)

HKJ said:


> ...
> It is the 189 and the other Fluke is the 289 that has replaced the 189 in the Fluke lineup.



So the 189 has replaced the "179" in the Fluke line up???
I notice you do have a 179.
The 189 is a new model? I have not seen this around???


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## Mr Happy (Jun 3, 2012)

peterharvey73 said:


> So the 189 has replaced the "179" in the Fluke line up???
> I notice you do have a 179.
> The 189 is a new model? I have not seen this around???



No, the 289 has replaced the 189 (which is now discontinued).

The 289 in my opinion is inferior in several respects, being bigger, more battery hungry, and having a more complicated user interface with lots of menus and stuff. Sometimes the march of progress is..."backwards" :shrug:


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## TedTheLed (Jun 3, 2012)

I got this fluke clamp meter to measure the maximum demand of motors when they start up, it seems to work..only $198...model 374 fluke clamp meter;

http://www.fluke.com/fluke/usen/electrical-test-tools/clamp-meters/fluke-374.htm?PID=70402

"..proprietary inrush measurement technology to filter out noise and capture motor starting current exactly as the circuit protection sees it.."


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## HKJ (Jun 3, 2012)

TedTheLed said:


> I got this fluke clamp meter to measure the maximum demand of motors when they start up, it seems to work..only $198...model 374 fluke clamp meter;
> 
> http://www.fluke.com/fluke/usen/electrical-test-tools/clamp-meters/fluke-374.htm?PID=70402
> 
> "..proprietary inrush measurement technology to filter out noise and capture motor starting current exactly as the circuit protection sees it.."




Very useful to measure large currents with, but it cannot measure power (But you can get clamp meters that can measure power).


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## TedTheLed (Jun 4, 2012)

whats power? pun not intended.

I mean if not volts x amps?


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## Russel (Jun 4, 2012)

TedTheLed said:


> whats power? pun not intended.
> 
> I mean if not volts x amps?



Power is volts X amps, but to measure power you have to measure current and voltage at the same time. That usually requires two meters.


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## HKJ (Jun 4, 2012)

Russel said:


> Power is volts X amps, but to measure power you have to measure current and voltage at the same time. That usually requires two meters.



That works fine for DC, but not for AC as I have written a couple of times.
With AC the current can be out of phase with the voltage, due to capacitors or inductors, there the "cos phi" is used to compensate. It can also be much more complex like a the rectifier in a power supply or a dimmer, in both cases the PF can be used to adjust the formula.
It is volt*current in all cases, but it need to be measured and multiplied a few 1000 times each second to get a usable result.


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## TedTheLed (Jun 5, 2012)

the fluke clamp does measure volts, you can attach test leads in the bottom..


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## HKJ (Jun 5, 2012)

TedTheLed said:


> the fluke clamp does measure volts, you can attach test leads in the bottom..



That can be used for DC power, not for AC power, except if the clamp has a power(Watt) range (Like the Fluke 345).


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## Russel (Jun 5, 2012)

HKJ said:


> That works fine for DC, but not for AC as I have written a couple of times.
> [...]



I stand corrected. I was thinking in terms of DC, my bad. Thank you for explaining.


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## TedTheLed (Jun 5, 2012)

HKJ said:


> That can be used for DC power, not for AC power, except if the clamp has a power(Watt) range (Like the Fluke 345).



the test leads sockets arent marked ac or dc, you choose which you want to measure with the dial..ac or dc...

"...Be ready for anything
The Fluke 374 True-rms Clamp Metercan read up to 600 V and 600 A in both ac and dc modes. Additionally, the 374 is compatible with the new iFlex flexible current probe (sold separately), which increases the measurement range to 2500 A ac and provides increased display flexibility, and improved wire access.
Safety Conformance..
EN/IEC 61010-1:2001; 1000V CAT III, 600V CAT IV
..."


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## HKJ (Jun 5, 2012)

TedTheLed said:


> the test leads sockets arent marked ac or dc, you choose which you want to measure with the dial..ac or dc...



Yes, of course, you can measure both AC and DC volts and the clamp can measure AC and DC ampere, but it cannot measure power and you cannot use the ampere*voltage to calculate power for AC (Except in a few special cases).


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## TEEJ (Jun 5, 2012)

Would the "Power" that results from whatever phasing is present be different if the phasing was different?

In other words, if I call the "power" the work that the motor can do at a given moment in time, for example...is it more "powerful" when the phases are, um, in phase?



For magnetic field interference for example, we sometimes make the "lobes" of the field cancel each other out, by adjusting phasing...to reduce the size of the magnetic fields.


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## HKJ (Jun 5, 2012)

TEEJ said:


> Would the "Power" that results from whatever phasing is present be different if the phasing was different?.



Yes, if the phase is 0 the power is 100%, but if the phase is 90 degree the power is 0. For anything in between use cos(phase) (This is called cos phi).

The problem with out of phase current is that you still have current and this gives losses in the cables and transformers.


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## TedTheLed (Jun 5, 2012)

...so, um, when I measure an inrush of 52 amps on a 110 volt system, I cant assume the motor wants 5720 watts, at startup..?


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## HKJ (Jun 5, 2012)

TedTheLed said:


> ...so, um, when I measure an inrush of 52 amps on a 110 volt system, I cant assume the motor wants 5720 watts, at startup..?




I am not an expert on motors but I believe it will be significantly below.
The motor is probably marked with the cos phi factor that is used to calculate power when it is running at normal load. There might also be a capacitor to compensate for the cos phi factor.


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## TedTheLed (Jun 5, 2012)

ya lost me there..but let me make sure Im being clear; I do want to know the absolute maximum the motor requires if even for a fraction of a second, since, for one thing, inverters shut down the moment a half an amp over maximum amps is reached.. so even if say, a pump on an inverter runs at 900 watts, when it is switched on it will require , accordimg to the fluke374, the 5720 watts from the inverter..and if the inverter maximum is 50 amps, it will shut down..


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## TEEJ (Jun 5, 2012)

HKJ said:


> Yes, if the phase is 0 the power is 100%, but if the phase is 180 degree the power is 0. For anything in between use cos(phase) (This is called cos phi).
> 
> The problem with out of phase current is that you still have current and this gives losses in the cables and transformers.




OK, good, I was making sure we were on the same page. I deal with the environmental/safety issues not the power USE issues normally.


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## desirider (Jun 5, 2012)

HKJ said:


> Yes, if the phase is 0 the power is 100%, but if the phase is 180 degree the power is 0. For anything in between use cos(phase) (This is called cos phi).
> 
> The problem with out of phase current is that you still have current and this gives losses in the cables and transformers.



I think net power is zero when phase is +/-90 degrees. When phase is 180 degrees it means negative power (cos(180)= -1), i.e. power is consistently flowing back into the supply. I am not sure when this could happen. Note that this is different from the reverse power flow due to an inductor or capacitor. In the case of pure inductors and capacitors power flows in both directions in _every_ cycle so it cancels out over a full cycle. But in the case of 180 phase shift between V & I power is flowing in one direction - back into the source. This may be physically impossible


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## TEEJ (Jun 5, 2012)

Again, my experience is with the resultant magnetic fields, which for a three phase system, produce "lobes" perpendicular to the direction of current flow at 120º angles...when its all lined up nice an purdy. 



(I think of it as a "3-Leaved Clover" shape)



When the phases are shifted to cancel out the magnetic fields, that should imply a reduction in involved induced electrical field influences as well....as it essentially rotates the lobes to new positions.

This in turn means the juice is somewhere ELSE than where it started, typically clocked proportionally, etc.


IE: MY involvement might be triggered if its NOT starting at 120º apart.


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## HKJ (Jun 5, 2012)

TedTheLed said:


> ya lost me there..but let me make sure Im being clear; I do want to know the absolute maximum the motor requires if even for a fraction of a second, since, for one thing, inverters shut down the moment a half an amp over maximum amps is reached.. so even if say, a pump on an inverter runs at 900 watts, when it is switched on it will require , accordimg to the fluke374, the 5720 watts from the inverter..and if the inverter maximum is 50 amps, it will shut down..



The inverter is probably measuring the current, not the power.
When working with AC power, there is usual 3 different values:
W, Watt, the real power
VA, Volt*Ampere, i.e. the volt*ampere value
VAR, Reactive power, this is the difference between W and VA



desirider said:


> I think net power is zero when phase is +/-90 degrees.



Yes, it was a typo.


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## jmac30 (Jun 6, 2012)

TedTheLed said:


> ya lost me there..but let me make sure Im being clear; I do want to know the absolute maximum the motor requires if even for a fraction of a second, since, for one thing, inverters shut down the moment a half an amp over maximum amps is reached.. so even if say, a pump on an inverter runs at 900 watts, when it is switched on it will require , accordimg to the fluke374, the 5720 watts from the inverter..and if the inverter maximum is 50 amps, it will shut down..



Ted your inrush/startup is going to be much higher than running current and can depend on too many factors to give you a specific answer - however taking an amperage reading with a clamp meter set to record your Max values should get you close enough to calculate the value you are looking for although I have to say I probably wouldn't recommend running a heavy motor load with a 12 volt DC inverter which is what it sounds like you are asking about good luck and post what exactly you are looking to do for more info

Sent from my Nexus S 4G using Tapatalk 2


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## TedTheLed (Jun 6, 2012)

thanks..in fact..I hope you like surprises; the example I was describing is the real live case here, or was..

had a half horsepower motor in the well. a Trace inverter that shut down over 50 amps, it is 2.5 kw continuous..so it can double that for peaks. Much measuring went on at first due to Trace shutting down every other time or every third time the pump was turned on..big mystery..
Until I called the very nice lady at the Grundfos company who told me, ta da, the pump I had could draw a max of 51 3/4 (or some similarly close number to 50) ! So, sometimes, the pump drew 50,
sometimes a fraction more and the digital machine went ding and shut down..

Problem solved by pulling up the pump and installing a Grundfos 1/3 hp instead..they dont makes those anymore, so I bought an extra..


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## HKJ (Jun 6, 2012)

TedTheLed said:


> thanks..in fact..I hope you like surprises; the example I was describing is the real live case here, or was..



Not surprised about that, only double up in starting current is not bad.


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## TedTheLed (Jun 6, 2012)

it more than doubled, though, a 1/2 hp grundfos pump with a 4 or 500 watt motor was taking 5,000 watts plus at start up..


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## jmac30 (Jun 6, 2012)

Even so that is really pretty good, by NEC and design handbooks you can have up to 25x (i cant remember exactly and have packed my codebooks away for a move)motor running current it makes for interesting sizing of fuses and over current protection devices -glad to hear you were able to solve it easily


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## HighlanderNorth (Jun 6, 2012)

Empath said:


> The amount of worked performed by a circuit today was 100 watts.
> 
> It did it with 100 amps. It didn't take much force since it had so many amps. It only took 1 volt.
> 
> ...





I know this is a VERY old post, but I must correct one major wrong in the analogy above^. It has nothing to do with electricity though, but I can tell you the analogy is unlikely or impossible, because no labor union is going to negotiate to do MORE work!


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## qwibbles (Jun 7, 2012)

Motor Manufacturers are always trying to improve the efficiency of their motors (Especially in modern times). One way of doing this for standard AC motors is to reduce the winding losses due to resistance. Doing this however increases the inrush current due to the lower resistance of the windings. During this inrush there is no established magnetic field in the windings etc of the motor, once the motor is running the changing magnetic field itself acts as a resistance to current flow (Changing magnetic fields) and reduces the maximum current that can flow. For this reason motors are normally protected by circuit breakers that are able to take 10-50 times (I don't know the exact range of the top of my head) of there normal rated current. For larger motors, due to this large inrush current, the number of times you can switch them on/off in a given period is limited to prevent the motor overheating. (Or extra cooling (Bigger Fans) are fittted). I believe the inrush current will get larger as motor efficiencies increase (There are methods of limiting the inrush but I will ignore those for the moment).


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## TedTheLed (Jun 10, 2012)

well that certainly would be appreciated information from here..solar people spend lots of time and money looking for motors that start and run easily with a smaller inverter..the well pump dictates the smallest inverter possible here, and that equals about a thousand dollars more for an inverter a thosand watts bigger than it would otherwise need to be, just to provide 5000 watts for a fraction of a second once a day...!


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## Mr Happy (Jun 10, 2012)

TedTheLed said:


> well that certainly would be appreciated information from here..solar people spend lots of time and money looking for motors that start and run easily with a smaller inverter..the well pump dictates the smallest inverter possible here, and that equals about a thousand dollars more for an inverter a thosand watts bigger than it would otherwise need to be, just to provide 5000 watts for a fraction of a second once a day...!



The answer would be a device that limits the turn on current to the motor and starts it up slowly. Big industrial motors (like traction motors on trains) have motor drive electronics that do this in a controlled way, but small motors (1/2 hp is tiny) are generally assumed to have no need for such sophistication.

0.5 hp is 425 W, so assuming a 120 V supply the motor running current would be about 4 A. To get 50 A at switch on indicates an inrush current ratio of over 10:1 at start up.

I suspect some device that limits the start up current, for example by putting a 1 kW halogen lamp in series with the motor while it starts up and then switching it out once the motor is turning, might do the trick. But I have not tried such a thing and don't know how well it would work with that exact motor and application.


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## TedTheLed (Jun 10, 2012)

if the bulb is in series with the motor then wont the circuit be broken when it is unscrewed?


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## Mr Happy (Jun 10, 2012)

It wouldn't be unscrewed, there would be a relay that bypasses the bulb a second or two after switch on. It would need a small bit of electronics to control the relay with a short time delay.

There are simple devices called NTC thermistors that are designed to provide inrush current limiting. Here is a link to a note about them:

http://www.epcos.com/web/generator/...operty=Data__en.pdf;/PDF_Applicationnotes.pdf

(I hope the link works.)


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## BoarHunter (Jun 11, 2012)

It would be simpler to explain using complex numbers, integral calculus and differential calculus, vectors etc... 

Inrush current is an issue not only with motors but many electronic apparatus. It is why in case of power outage, so many circuit breakers will trip when power comes back provided the equipments have not been turned off.


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## jmac30 (Jun 11, 2012)

Ted - they make soft starts and similar devices for motor inrush mitigation - unfortunately they are typically a 3 phase application - this seems like a pretty good niche to be involved in - hmmm

Sent from my Nexus S 4G using Tapatalk 2


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## Wrend (Jun 11, 2012)

Sounds like a mix up of electron inertia and a stalled/not yet started motor?


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## truerock (Mar 23, 2013)

Nerd said:


> Lemme have the honour of taking up the challenge. Let's take electricity as water. Wire as garden hose.
> 
> Voltage is amount of pressure. So more voltage, more pressure, water shoot out of hose faster.
> 
> ...



I humbly suggest this analogy (although common) is incorrect. The diameter of the hose is analogous to ohms (i.e. resistance). Amps is analogous to the weight of the liquid traveling through the hose (e.g. water versus, crude oil). Watts would be analogous to pounds per minute of liquid traveling through the hose. At least the volts being analogous to pressure was correct


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## davevolt (Sep 26, 2013)

Nerd said:


> Lemme have the honour of taking up the challenge. Let's take electricity as water. Wire as garden hose.
> 
> Voltage is amount of pressure. So more voltage, more pressure, water shoot out of hose faster.
> 
> ...



How does power factor affect this? For example, my aquarium pumps run about 80 watts total, and the lights add an additional 350. On my kill-a-watt power meter it is showing 420 watts consumption, which is about what adding the 3 pumps, the light wattage and ballast loads would equal, but on the current meter, it is drawing a whopping 11.5 amps of current, and I believe it, because when the lights are running the 15 amp rated power bar cord gets warm to the touch as does the electrical outlet, so it is drawing a huge amount of current. The power factor is about 0.35 which is an indication of the highly inductive load that the 3 pumps and 2 magnetic metal halide ballasts are pulling.


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## mattheww50 (Sep 26, 2013)

Energy is the integral over time of Volts x Amps. Power (Watts) is the instanteous product of Voltage x Current. If the Volts and Amps are not exactly in phase, then Volts x Amps does not equal energy per unit of time or Watts. Most AC equipment is in fact rated for something called Volt-Amps or simply VA, which really is Volts x Amps. The power factor is the cosine of the phase angle between Voltage and Current. If the power factor is 1, then VA also equals Watts or Power

As the phase angle approaches 90 degrees, the actual power consumed approaches zero. This in fact makes you very unpopular with the electric utility because regardless of Actual Energy you consume, the formula for loss in transmission is still I^2 x R. So at very low power factors you force the supplier to incur large losses in the tranmission system, while the watt hour meter that the bill comes from hardly moves at all. For Commercial users, there is in fact a substantial surchrage for lower factor consumption (generally below .8). In your example the power factor would be ~3/11.5, so the phase angle betweem the voltage and the current would be arc cosine (3/11.5). If you were a commercial user, you'd pay a substantial surcharge for such a low power factor.


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## StorminMatt (Sep 28, 2013)

Nerd said:


> To make it as unconfusing as possible, take your normal everyday garden hose: You pinch it, you get higher voltage, you change hose to bigger hose, more water can flow, but that doesn't = higher voltage because water may flow slower instead.



Interestingly, what this paragraph best describes is internal resistance. Just as an electrical circuit has resistance associated with the power source (like a battery) and the load (like a light bulb), a fluid system has flow resistance associated with both the source (plumbing, valves, etc) and the load (like a garden hose, sprinkler, etc). When you have a large, free flowing hose connected to a faucet, its resistance is low compared to the resistance of everything else in the system. So most of the pressure drop occurs upstream from the hose (ie in the piping, valves, etc). But when you pinch down the hose, you increase the load resistance. This reduces flow through the system, which reduces the pressure drop before the pinch in the hose. This is why pinching a hose causes pressure to increase.

Likewise, with an electric circuit, if the load resistance is low compared to the source resistance, the voltage delivered to the load is going to be low. A well known example of this around here would be trying to run an ROP 3854H bulb on six alkaline D batteries in a 6D Mag. Because the ROP Hi bulb trys to draw so much current, it brings the voltage of the high resistance alkaline batteries down. But if we instead use a 6D Magnum Star bulb, we are effectively pinching down the hose. By increasing the load resistance, we bring down the current. And because there is less voltage drop through the internal resistance of the batteries, the output voltage of the batteries increases.

One thing, though. Flow will not decrease with decreasing resistance in either an electrical or fluid system. The maximum flow ALWAYS occurs when the load resistance is zero. Keep in mind that, in this situation, the pressure (fluid system) or voltage (electrical system) across the load is zero at this point.


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## havequick (Oct 12, 2013)

StorminMatt said:


> One thing, though. Flow will not decrease with decreasing resistance in either an electrical or fluid system. The maximum flow ALWAYS occurs when the load resistance is zero. Keep in mind that, in this situation, the pressure (fluid system) or voltage (electrical system) across the load is zero at this point.



In any generic electrical system, maximum power *transmission* occurs only when the source and the load are at matched impedances, not when the load is zero. It is an important but critical distinction because no source will have a zero internal resistance, even for low frequency systems.


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## StorminMatt (Oct 12, 2013)

havequick said:


> In any generic electrical system, maximum power *transmission* occurs only when the source and the load are at matched impedances, not when the load is zero. It is an important but critical distinction because no source will have a zero internal resistance, even for low frequency systems.



Maximum power TRANSMISSION will always occur when the source impedance is equal to the load impedance. But maximum power OUTPUT from a source will always occur when load resistance is zero. This is kind of the jist of what I was saying. Specifically, current will be greatest if the load resistance is zero (ie a dead short). However, as I said, voltage will also obviously be zero across a dead short, resulting in all power being transferred to the internal resistance, and being dissipated as heat by the source.


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## Mr Happy (Oct 13, 2013)

StorminMatt said:


> Maximum power TRANSMISSION will always occur when the source impedance is equal to the load impedance. But maximum power OUTPUT from a source will always occur when load resistance is zero. This is kind of the jist of what I was saying. Specifically, current will be greatest if the load resistance is zero (ie a dead short). However, as I said, voltage will also obviously be zero across a dead short, resulting in all power being transferred to the internal resistance, and being dissipated as heat by the source.



I think you've missed something crucial there.

You suggest maximum power output from a source will occur when the load resistance is zero.

But then you go on to observe that when the load resistance is zero there will be no power in the load and all power is dissipated in the source. This is another way of saying the power output from the source is zero. All of the power is turned to heat inside the source and no power leaves the output terminals.

Now if you say that maximum power _consumption_ occurs within the source when the load resistance is zero, then that is something different.

But power output by definition means power coming out of the output terminals.


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## mattheww50 (Oct 13, 2013)

Mr Happy said:


> I think you've missed something crucial there.
> 
> You suggest maximum power output from a source will occur when the load resistance is zero.
> 
> ...



Not quite true, in either AC systems, or systems that involve an energy storage component. The reality is zero resistance loads actually do exist, the catch is they are no zero impedance loads.
For example a Magnetic Resonance Imaging Scanner (MRI) contains a true zero resistance device, it is a super conducting magnetic, and there is a lot of energy stored in a 3T magnetic field. As a super conducting magnetic you can view it as an inductor with an infinitely high Q, since it is truly a lossless device. If you connected such a load to an AC source, the source could be producing maximum Volt-Amps, but the Power factor would be Zero, so no power is actually consumed, i.e. the source could in fact be zero resistance. 

This incidentlally is an extreme example of why AC equipment is rated in Volt-Amps, not watts.


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