I'm trying to build one and learn a few things in the process. When i try to google it I get the wrong question answered (like how to read a electric meter, etc.). AC just looks like a sine wave on an oscilloscope so what characteristics of that sine wave are measured to determine wattage used? If that's not what is looked at, then what is?
Secondly, if a 1:1 transformer is used between the meter and the load, would the meter still read correctly?
Hi
There's quite a discussion on the theory and mathematics behind AC metering over at OpenEnergyMonitor.org. They've also wrapped up the theory into an Arduino library for ease of use.
Cheers ! Geoff
Watts = Voltage * Current, yes?
So if you know the voltage, all you have to do is meaure the current.
If you have a low-impedance transformer that all your current goes thru with a small secondary winding that drives a motor with spins a dial/set of dials, the current used can be displayed, and multiplied by the known voltage to provide the wattage.
What about the second question?
David82:
I'm trying to build one and learn a few things in the process. When i try to google it I get the wrong question answered (like how to read a electric meter, etc.). AC just looks like a sine wave on an oscilloscope so what characteristics of that sine wave are measured to determine wattage used? If that's not what is looked at, then what is?Standard AC voltage and current units of measurements are given in equivalent DC RMS values such that if your were to power a 120 ohm resistor from both a 120 vac power source and another time with a 120 vdc battery, both are said to create 1 amp of RMS current and the resistor is dissipating 1 RMS watt of power. AC voltage can be measured in other units like average voltage, peak voltage, peak to peak voltage, all derived from the same indentically sized AC voltage wave form.
The measurement of true AC power consumption (as measured by your billing meter) involve other additional factors like Power Factor (phase angle difference between voltage and current for loads other then pure resistance) and dealing with non-sine wave current flows ( the AC voltage provided to you is always in sine wave form, but some of your loads don't draw current as a pure sine wave and must be converted to true RMS current to obtain true power consumption). But before advancing you first should nail down voltage/current/wattage measurements as they apply to AC and DC circuits and what the units of measurement really mean. This seems to be a good article on the subject: http://www.allaboutcircuits.com/vol_2/chpt_1/3.html
Secondly, if a 1:1 transformer is used between the meter and the load, would the meter still read correctly?
Yes, the 1:1 ratio tells you the voltage and current levels with be equal between the primary and secondary minus a few percentage power loss in the transformer core in the form of heating of the iron core due to eddy currents.
Lefty
We studied phase angle in my college physics 2 class. Is the phase angle what is used to measure current?
When using a 1:1 transformer, I don't understand how the meter can still tell how much current is used. The wires providing power aren't actually connected to the load in any way.
David82:
We studied phase angle in my college physics 2 class. Is the phase angle what is used to measure current?No, phase angle is the relationship between voltage and current at an given instant of time and if not 0 degrees then simply multiplying voltage X current won't give you the true RMS power consumption. The phase angle must be measured so it's effect can be calculated for true power consumption as in billing rate.
When using a 1:1 transformer, I don't understand how the meter can still tell how much current is used. The wires providing power aren't actually connected to the load in any way.
Not directly connected but effectively connected by the transfer of power via constantly changing magnetic coupling from primary to secondary windings, the iron core of the transformer makes this a very efficient transfer of power. This only works for AC as DC will not pass power through a transformer as there is no changing magnetic field as in AC. AC power transfer is effectively constant (minus small transformer losses) such if the step up voltage ratio is 1:2 the current ratio transfer will be 2:1 making the actual power transferred constant.
As pointed out above, for an extended well researched answer go to openenergymonitor.org.
But, in direct answer to your question: A 1:1 transformer won't do it for you. As you pointed out, the current doesn't go through it. You have to measure the voltage and the current for many intervals, and square, sum, root, then average them; that's what RMS (root mean square) means. Then, you have to consider the errors introduced by phase angle of the various components you used and the offset due to calibration of the same components. For us hobbyist, current transformers measure the current and an isolation transformer is used to measure the voltage. We sample the voltage and current any times a second and do the calculations on an arduino to give use a value that varies over time and usage. Those devices can be searched for on Google.
This is what CrossRoads was hinting at below. The entire explanation is quite involved and more than I should go into here. If you want an explanation of power meters as the power companies use them, go to:
Over the years there have been many techniques used ranging from full analog relying on induced currents on an aluminum disk to high speed digital sampling. Take your pick from the article. Some of the companies that make the power meters have descriptions and modes of operation online for you to look at; look at your own meter and try to find it online. You might get lucky.
Secondly, if a 1:1 transformer is used between the meter and the load, would the meter still read correctly?
If you put the transformer between your electric meter and your referigerator, of course the meter is still going to measure the power consumption...
AC just looks like a sine wave on an oscilloscope so what characteristics of that sine wave are measured to determine wattage used?
If you have a resistive load, like a light bulb, the current and voltage are in-phase. So, you can simply measure the RMS current and the RMS voltage, and multiply.
With inductive or capacitive loads, the current and voltage can be out-of-phase by some amount (up to 90 degrees). In that case you have to measure the voltage and current at the same instant and calculate the power for that instant, or take many measurements or measure continuously to get an RMS power measurement. (I'm sure you use some calculus in your physics class, and you might remember how to multiply two sine waves with a known phase-angle difference...)
If you call an electrician and ask him to measure the power your refigerator is consuming, he might not have a power meter handy*, so he''d probably break the circuit and insert his multi-meter in series to measure RMS current. Then, he'd multiply by the known line-voltage to calculate the power. That wouldn't be 100% accurate, since the compressor motor is somewhat inductive, but it would be a reasonable approximation.
If you wanted to do someting similar with the Arduino, it would be easier to measure the peak current. Since you know it's a sine wave, you can multiply by 0.707 to get the RMS current. And, since you also know the (approximate) line voltage, you can calculate the power.
If it's for a physics class or for the power company, it's very important to measure the true-accurate RMS power consumption. For casual/hobby use you may be able to take some shortcuts. 
*Power meters are somewhat rare, but I assume most electricians now have a [u]Kill-A-Watt[/u] in their truck.
One of these devices does all the hard work for you.
Measuring AC power accurately isnt a simple thing to do,so best not to reinvent the wheel.
This is for a college project. We are supposed to figure out a way to power a load without the meter registering it 
Some students were thinking you could draw a really small amount that maybe wouldn't register. Another idea was to suck up a huge amount faster than the sampling rate and dissapate that slowly to power the load. Another idea was the 1:1 transformer but that I guess wouldn't work. Any ideas? The school allocated the funds for us to build it.
Not be able to measure it in what way, current or voltage? If it's voltage, you could max out (overload) the meter. Find out what the maximum voltage it can read then try to go beyond it. (Dangerous though) You could use a step-up transformer or you could get it to oscillate beyond what the meter can read.
David82:
This is for a college project. We are supposed to figure out a way to power a load without the meter registering it
Some students were thinking you could draw a really small amount that maybe wouldn't register. Another idea was to suck up a huge amount faster than the sampling rate and dissapate that slowly to power the load. Another idea was the 1:1 transformer but that I guess wouldn't work. Any ideas? The school allocated the funds for us to build it.
Simple attach/wire your load to the input side (upstream) of the metering point rather then output side (downstream).
Lefty
HazardsMind:
Not be able to measure it in what way, current or voltage?
current
retrolefty:
Simple attach/wire your load to the input side (upstream) of the metering point rather then output side (downstream).Lefty
The instructor joked about doing that. Not allowed. We have to power something, like a light bulb without the meter registering it.
Good one Lefty!
I can think of a couple of other possibilities off hand. If you only need a little power, just wrap some wires around one leg of the incoming before the meter and suck some power out inductively. The other one is to use something that has a very low power factor and returns everything back to the power company as reflected power. I used to do this with a fluorescent tube, just connect the ends without turning on the filaments then strike a spark with something else to get it to fire up and away you go, free light. You can also heat the filaments with a battery and get it to fire and disconnect the battery; big battery though.
A long, long time ago with completely different meters than we have today, we would use a really powerful magnet and retard the aluminum disk by putting the magnet right up against the meter's glass and holding it on with duct tape. This wouldn't stop the meter, but it would slow it down noticeably. Doesn't work with any of the new meters though.
You could also slip a bribe to the guy that reads the meter.
Did he specify what meter he'll be using?
About the only way I can think of doing this is to try and exploit the accuracy limitations that all
electronic electricity meters have.
Most modern electronic electricity meters are Class 1 meters which have a variety of accuracy specs
depending on how much energy they have measured.
A modern utility meter is Class 1 with a very tight requirement to be accurate to +/- 1% down to 5% of load.
If the Meter has a maximum current rating of 100 A, then the meter will be accurate within 1% down to
50 ma , so anything below this means that the meters accuracy is not defined.
If the supply voltage is 240 V , then a load of less than 12 watts will be below the meters minimum accuracy standard
and whilst the meter may measure this amount of power, you could most likley argue that the reading is not accurate.
Running a 10 mw led of such a meter would most likley not be recorded at all.
I've no practical experience of how electricity meters work, but I seem to remember when I was taught about power factors being told that these meters typically measure current, and that the mean square adjustment and integration over time was then used to calculate the transferred charge and hence energy. This was to explain why industrial consumers were concerned about power factor - because they are charged for current but get value from energy so they want the power factor to be as high as possible so they don't pay for energy they didn't use.
Given that these devices are specifically designed to measure current accurately over a wide range of conditions, I doubt that it will be easy to cheat them. You could demonstrate reducing the power factor so that the meter shows power that you didn't get, but there's no way (that I know of) to get a power factor greater than 1.
What scope do you have to interfere with the power supply coming in to the meter? Is it entirely under your control, are you allowed to use inductive coupling etc, or is it completely inaccessible to you? If you only need to demonstrate power going through the meter without being measured and the whole circuit is under your control, then you could try either using DC (I have no idea whether meters would detect that) of a very high frequency AC well outside the range the meter is designed to pick up.
PeterH:
....the whole circuit is under your control, then you could try either using DC (I have no idea whether meters would detect that) of a very high frequency AC well outside the range the meter is designed to pick up.
what do you mean by this?