Save power with the "power save 1200"?

[DRBUZZ0]

Just saw a commercial for this thingy: http://www.power-save.com/

Basically it connects to your electrical system and is supposed to "optimize" the power of a home or commerical electrical system and thereby reduce power consumption.

I don't have time to research it right now, but I'm very skeptical. From the page it sounds like it's some sort of impedance balancing unit or power conditioner. That's not entirely woo, but it claims results of up to 25% reduction in power consumption. That sounds fishy to me. That is, unless it's just a big resistor, which would certainly reduce power usage but also would have a tendency to dim all your lights and make motors grind a bit in appliences.

Anyone heard of this?

[JoeEllison]

It is just a voltage regulator, isn't it?

[Dilb]

They say right in the FAQ that it's a capacitor bank. The idea is that by increasing your power factor, the meter will read less. I'm not an electrical engineer, but as I understand things: fifty years ago this might have been an issue, but it shouldn't be a problem anymore. Improving your power factor does actually save energy, because it reduces the total current in the wires needed to deliver a certain amount of power, and less current means less energy lost due to resistance. Power companies are well aware of this fact.

Edit- Also, they don't claim to " results of up to 25% reduction in power consumption", but rather " save customers as much as 25% on their electric bill every month ". It's a savings on the bill, not necessarily in terms of actual energy.

[casebro]

Resistors don't save energy. What does not go through them, is given off as heat.

If capacitors could save 25%, then don't you think appliances would come with capacitors built in? Gosh, I just realizes: myu refrigerator motor has a capacitor. My TV is loaded with them. My PC power supply has several. I'll bet my fluorescent replacements for incandescent bulbs have capacitors in them.

I can only think of one thing in my house that doesn't have a capacitor. So how would adding a capacitor to my fart fan circuit save me 25%?

[RecoveringYuppy]

Resistors don't save energy.
If you put a resistor in series with your load you will cut down the energy consumed in the circuit, though possibly at the expense of proper operation of the appliance.

V=IR and P=R I^2 are the relevant equations.

[marting]

If you put a resistor in series with your load you will cut down the energy consumed in the circuit, though possibly at the expense of proper operation of the appliance.

V=IR and P=R I^2 are the relevant equations.


True for a resistive load. However that is not always the case. If you put a resistor in series with a device that uses regulated power such as a laptop supply, you will increase the overall power level.

[RecoveringYuppy]

marting: True. If the device is "smart" or "active" and changes it's resistance in response to the voltage applied then anything could happen.

If capacitors could save 25%, then don't you think appliances would come with capacitors built in?
A bit of googling seems to indicate that modern appliances do incorporate something like this already.

[technoextreme]

marting: True. If the device is "smart" or "active" and changes it's resistance in response to the voltage applied then anything could happen.


You mean a transistor?? That's the first time I've ever heard of a transistor be referred as an smart device. Active yes.
If capacitors could save 25%, then don't you think appliances would come with capacitors built in? Gosh, I just realizes: myu refrigerator motor has a capacitor. My TV is loaded with them. My PC power supply has several. I'll bet my fluorescent replacements for incandescent bulbs have capacitors in them.
Congratulations, you just cited five different appliances with capacitors that probably have five different purposes. In fact the only thing that probably has an relevance to this conversation is the refigerator motor which would have a pretty large inductive current surge.

[AgeGap]

It may balance capacitance and inductance. It may store low tariff power for use during high tariff times.

"Be a Power-Save Authorized Re-seller / Installer!" is setting off alarm bells.

Even if it isn't woo how long will it take to pay for itself?

[GWB/]Just send less e-mails to save on your power bills[dubya]

[RecoveringYuppy]

It may store low tariff power for use during high tariff times.
The capacitor capable of doing that amount of storage would be very expensive. This is just lowering the reactive load of the device which means the energy is being saved for a fraction of a second.

[technoextreme]

It may balance capacitance and inductance. It may store low tariff power for use during high tariff times.

"Be a Power-Save Authorized Re-seller / Installer!" is setting off alarm bells.

Even if it isn't woo how long will it take to pay for itself?

[GWB/]Just send less e-mails to save on your power bills[dubya]
Actually, I understand what it does now. In theory it sounds like they made something like regenerative braking. Don't know how feasible it is but that's what they claim.

[Hindmost]

It is just baloney....Power is still current times voltage...whatever you turn on is going to determine how much power you use. if you want to save on your bill turn off the A/C, refrigerator, freezer and toaster and hairdryer. (and the coffee pot if you can live without coffee--which is impossible for me.)

The utility actually sells energy, not electricity...kw-hrs are actually joules of energy since a kw is a joule per second.

http://en.wikipedia.org/wiki/Electric_power

Utilities don't charge for apparent power...only real power...utilities use enormous capacitor banks to change the power factor to as close to unity as possible--this reduces their capital cost of equipment as it reduces reactive power and circulating currents due to inductive loads.

glenn

[LongFuzzy]

If I understand the theory, if you have a lot of electric motors (as in industrial factory) then correcting the power factor will help. (Factories already know this and do it, as does the power company) In this case the "power-saver" is too small. If you live in a residence, and only have a few motors, that don't run full time, then the "power-saver" is useless. The power companies know all about power factors, and build the electrical meters to measure the actual energy you use.

[mhaze]

If I understand the theory, if you have a lot of electric motors (as in industrial factory) then correcting the power factor will help. (Factories already know this and do it, as does the power company) In this case the "power-saver" is too small. If you live in a residence, and only have a few motors, that don't run full time, then the "power-saver" is useless. The power companies know all about power factors, and build the electrical meters to measure the actual energy you use.

Don't forget that all it has to do to succeed is fool the electric company's meter on your house into thinking you use less power than you do. So the important thing to know is how that meter integrates the electrical flow used.

Which I don't know. Nonetheless, it is the way to understand whether this device could save 25% of a utility bill....

[marting]

Don't forget that all it has to do to succeed is fool the electric company's meter on your house into thinking you use less power than you do. So the important thing to know is how that meter integrates the electrical flow used.

Which I don't know. Nonetheless, it is the way to understand whether this device could save 25% of a utility bill....

Most every electric meter going into a house reads KWH (energy) and they aren't fooled. More advanced units may also measure time of day and power factor but these are not common in residences.

Banking capacitors in parallel will have multiple effects.

It will will tend to cancel net inductive currents. To the degree this reduces ohmic loss along the resistive path inside the house to the utility meter it will save energy. This is described (accurately) as R1 on page 4 of Mutlu and Rahman's "supporting" paper. However, the paper avoids quantitave examples. For reasons that are all too apparent, IMO.

It will also provide a capacitive current path that operates at all times. When household motors are off and other demand is low, this fixed current results is quite a low power factor. While this current is lower than that of the motors, it is now present at all times when the motors are off. This is the vast majority of the time for many appliances. During this time the utility meter will measure that energy deposited into "R1" as well as other losses, such as those in the capacitor itself. This is alluded to indirectly in the "ongoing investigation" section after the conclusion.

That said, there is absolutely nothing in the paper, which is largely elementary electrical theory, that supports a 25% savings (or reduction in energy). Ohmic wiring losses associated with a power factor of .8 vs 1.0 are generally very low to start with. Well under 5% unless the home has long runs of high gauge wire. These are the only area of potential, material savings.

I simply don't believe the FAQ's numbers of 15 to 20% typical savings. Depending on how frequent the motor load is, it might even be a net increase in energy consumed due to ohmic loss in the always active capacitor and associated circuits.

[Lensman]

From Wiki:-
The power factor of an AC electric power system is defined as the ratio of the real power to the apparent power, and is a number between 0 and 1. Real power is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power. Low-power-factor loads increase losses in a power distribution system and result in increased energy costs.

It's been a few decades since I did any power factor correction problems, that's why I had to resort to Wikipedia.

A bank of capacitors CAN improve the actual power usage in your home if you use a lot of inductive devices, but they normally have some compensation built in (the capacitors mentioned in one of the posts above) - what they will NOT do is store power for use later - capacitors can only store DC electricity - they CANNOT store AC to be released & used some hours later.

The full Wiki article is Here (http://en.wikipedia.org/wiki/Power_factor)

[mhaze]

Most every electric meter going into a house reads KWH (energy) and they aren't fooled. More advanced units may also measure time of day and power factor but these are not common in residences.

Banking capacitors in parallel will have multiple effects.

It will will tend to cancel net inductive currents. To the degree this reduces ohmic loss along the resistive path inside the house to the utility meter it will save energy. This is described (accurately) as R1 on page 4 of Mutlu and Rahman's "supporting" paper. However, the paper avoids quantitave examples. For reasons that are all too apparent, IMO.

It will also provide a capacitive current path that operates at all times. When household motors are off and other demand is low, this fixed current results is quite a low power factor. While this current is lower than that of the motors, it is now present at all times when the motors are off. This is the vast majority of the time for many appliances. During this time the utility meter will measure that energy deposited into "R1" as well as other losses, such as those in the capacitor itself. This is alluded to indirectly in the "ongoing investigation" section after the conclusion.

That said, there is absolutely nothing in the paper, which is largely elementary electrical theory, that supports a 25% savings (or reduction in energy). Ohmic wiring losses associated with a power factor of .8 vs 1.0 are generally very low to start with. Well under 5% unless the home has long runs of high gauge wire. These are the only area of potential, material savings.

I simply don't believe the FAQ's numbers of 15 to 20% typical savings. Depending on how frequent the motor load is, it might even be a net increase in energy consumed due to ohmic loss in the always active capacitor and associated circuits.

If you draw a sine wave and then consider integrating it over time, you'd have to make a really big chop in that sine wave before the integration was 15-20% less. Without taking a scope out to the house's AC input, I already know that it is going to look exactly like a sine wave (anybody with a scope has from time to time looked at the AC sine wave). Transient evenets, like when an AC compressor 5 hp motor kicks in, are not going to substantially affect total current draw.

There would be no other way to get to 15-20% other than "fooling the meter". I'm agreeing with you, just trying to formulate what could be the only plausible way they could do what they say they do...

[marting]

There would be no other way to get to 15-20% other than "fooling the meter". I'm agreeing with you, just trying to formulate what could be the only plausible way they could do what they say they do...


I believe the 15-20% claim in the FAQ is just plain false.

[bjb]

Here's a possible reason for their claim. They are comparing someone electricity bill from last year to this year's bill, and it shows a dramatic drop in energy use. However, at my house, our bill for July dropped from $490 to about $90. That's because last year was unusually hot and we ran the A/C day and night. This year the weather was much better and we didn't run it nearly as much. That 25% number could be true but their measurement method is very unfair.

[Fnord]

It is just baloney....Power is still current times voltage...

Power is actually Current times voltage times the cosine of the phase angle between the two, or...

P = E x I x COSΘ

If the phase angle is zero, which is the case in a purely resistive circuit, then COSΘ will be 1, and P will indeed equal E x I.

If the phase angle is 45°, then COSΘ will be 0.707, which is the case when the reactance (capacitive or inductive) has the same scalar value as the resistance. For example, if R = 1000Ω, and XL = 1000Ω, then ZT = 1414Ω, and COSΘ = 0.707. Thus, if E = 120VAC and ZT = 1414, then I = 0.085AAC and P = 120 x 0.085 x 0.707 = 7.21W, as would be the case when using an AC motor with a high winding resistance.

If ZT were a purely resistive load (or if the voltage supply was DC), then the power would instead be P = 120 x 0.085 x 1 = 10.2W.

QED: Increasing the reactance of a load would cause a reduction in the apparent power.

... no more brain ... out of think ... need lie-down ...

[tracer]

They should have called it the Power Saver 2000.

It would be, like, 800 better.

[marting]

Actually, the original statement that power=voltage*current is correct. When the waveforms are non static, one integrates power over time to get energy. The simplifications you make assume a mix of reactive and resistive loads to an AC source. They produce an average of the power over complete cycles.

Power is actually Current times voltage times the cosine of the phase angle between the two, or...

P = E x I x COSΘ

If the phase angle is zero, which is the case in a purely resistive circuit, then COSΘ will be 1, and P will indeed equal E x I.

If the phase angle is 45°, then COSΘ will be 0.707, which is the case when the reactance (capacitive or inductive) has the same scalar value as the resistance. For example, if R = 1000Ω, and XL = 1000Ω, then ZT = 1414Ω, and COSΘ = 0.707. Thus, if E = 120VAC and ZT = 1414, then I = 0.085AAC and P = 120 x 0.085 x 0.707 = 7.21W, as would be the case when using an AC motor with a high winding resistance.

If ZT were a purely resistive load (or if the voltage supply was DC), then the power would instead be P = 120 x 0.085 x 1 = 10.2W.

QED: Increasing the reactance of a load would cause a reduction in the apparent power.

For a resistive and reactive component in series but not if they are in parallel. However, it's actually beside the point.

The capacitor bank in this product will tend to cancel the inductive reactive component to some degree but will have little effect on the appliance's power consumption. Specifically, it does not materially change the delivered voltage and hence cannot materially change the consumed power. What difference there is is the miniscule amount due to the decrease in current from the circuit-breaker box (the caps and appliance runs into) and the electric utility meter. That just might account for a few tenths of a volt over a typically short, thick, wire.

The reason factories use pf compensation is mostly becaused they get charged a penalty by the utility if the pf gets too low.

Oh, and the utility meter responds to actual power and integrates it mechanically over time into energy so a pf correction bank has little effect.

[marting]

For a resistive and reactive component in series but not if they are in parallel.


Retracted. The apparent power is decreased.

Rest of comment stands

[balrog666]

They should have called it the Power Saver 2000.

It would be, like, 800 better.

Thanks for the laugh!

:th:

[Fnord]

There's a rather decent Second-Year tutorial on impedances on the other end of THIS LINK (http://www.faqs.org/docs/electric/AC/AC_5.html). It deals with complex numbers - values expressed with both real and imaginary numbers - and may be somewhat difficult those who have not had First-Year algebra. Maybe not.

[marting]

There's a rather decent Second-Year tutorial on impedances on the other end of THIS LINK (http://www.faqs.org/docs/electric/AC/AC_5.html). It deals with complex numbers - values expressed with both real and imaginary numbers - and may be somewhat difficult those who have not had First-Year algebra. Maybe not.

Assuming most people here understand complex numbers and given the reasonable simplification we are dealing with linear networks and a 60hz source, what exactly are your points re the PF correction device? My points were that it has little impact on measured energy usage. Are you making a different argument? If so could you outline how?

[Fnord]

... what exactly are your points re the PF correction device?

This particular one ... useful, but only if all of your motorized appliances are run by uncompensated electrical motors. Even then, without a PF meter on your house mains, you would never know if you have too much or too little PF compensation.

My points were that it has little impact on measured energy usage. Are you making a different argument?

No essential difference from your argument, except to say that there may be some conditions where it might be useful, but that the average homeowners, having only the manufacturer's assurances to go by, would likely waste their money (at best), or burn their house down and/or kill people (at worst).

Most modern-day motorized appliances already have some PF compensation built in (usually a capacitor in parallel with the motor windings and the mains connection) or use digitally-driven multi-phase stepper motors. Others have a starter capacitor in series with one side of the mains, and connected to a "Phase Tap" to make sure that the motor starts its rotation in the right direction, and that then provides some PF correction when the motor comes up to speed. Otherwise, there is little that the average homeowners can do.

Yes ... I know about those PF correction devices they sell at the big-box hardware stores. They're useful, but only on older appliances, or in places that run a lot of uncompensated ac motors.

[Elind]

I think these have been around for a long time. As has already been explained, a motor or other device consumes X current when it runs, according to its design. That cannot be externally reduced, nor can its power usage.

However, electric motors can be controlled to start up more efficiently than they typically do and they waste more power when starting than they do when running at design speed. Large motors have startup circuits to both protect the motor from overheating, or circuit breakers from tripping when they are essentially a short circuit before they start to move.

I believe these devices are supposed to reduce the startup surge when a motor starts, thereby saving some power. For a fridge that goes on/off frequently, it might make a difference, but how much I don't know. It would probably be much more cost effective to pay a bit more for the higher efficiency fridge, which will have more efficient motors as well as insulation, than a device like this for the old clunker.

ETA. The post above seems to say essentially the same thing, which I didn't read to before posting, but the more the better.

[marting]

This particular one ... useful, but only if all of your motorized appliances are run by uncompensated electrical motors. Even then, without a PF meter on your house mains, you would never know if you have too much or too little PF compensation.



No essential difference from your argument, except to say that there may be some conditions where it might be useful, but that the average homeowners, having only the manufacturer's assurances to go by, would likely waste their money (at best), or burn their house down and/or kill people (at worst).

Most modern-day motorized appliances already have some PF compensation built in (usually a capacitor in parallel with the motor windings and the mains connection) or use digitally-driven multi-phase stepper motors. Others have a starter capacitor in series with one side of the mains, and connected to a "Phase Tap" to make sure that the motor starts its rotation in the right direction, and that then provides some PF correction when the motor comes up to speed. Otherwise, there is little that the average homeowners can do.

Yes ... I know about those PF correction devices they sell at the big-box hardware stores. They're useful, but only on older appliances, or in places that run a lot of uncompensated ac motors.

I guess the only place we differ is that I can't even come up with a hypothetical with an uncomped motor where the device would make a material difference, let alone a 15 to 20% reduction in metered power. Even where the apparent power was 40% higher due to a pf of .7 the decrease in actual power (as metered on the house) by an attached comp would be quite small (much less than 5%).

I think the products claims are wildly exaggerated.

[Kopji]

They should have called it the Power Saver 2000.

It would be, like, 800 better.


Our new demonstration video shows how our residential Power-Save 1200 and commercial / industrial Power-Save 3400 work within their respective applications.


You should like, apply for the million bucks. :)

As I recall, most of the large motors in my house already have capacitors on them and so do yours.

[DRBUZZ0]

I can see how the theory kinda - sorta works, but I don't have that many inductive motors, now that I think of it, aside from the air conditioner... which I suppose is a pretty large power consumer. Also the fridge. Maybe the furnace blower too. But I recon most of the power I use is probably from non-motor stuff like lighting, electronics and such.

I would think the big inductive motors ought to have some motor-start caps on them and all the other stuff to help with the inductive loads.

On top of that the power in my area is pretty damn good. Nice sign wave and keeps a good 120 volts RMS steady just about all the time.

So...

I could see how this would work okay if you had some big inductive motors coming on and off all the time, on single phase AC without capacitors or anything.

But for the standard home? I would doubt it could make much difference.

[marting]

I could see how this would work okay if you had some big inductive motors coming on and off all the time, on single phase AC without capacitors or anything.

But for the standard home? I would doubt it could make much difference.

Even there I don't see much impact. The AC load during turn on is a decaying sine current that is highly reactive and starts at 5 to 20x the cont max load. As the motor rotation increases, the reactive load decreases over perhaps 10 cycles or so. The problem is the capacitor bank counters only a small fraction of this since it's contribution is a continuous, nearly constant, reactive current designed to counter the normal running loads. Net effect on measured energy consumed, essentially nada. Easy enough to model with Spice but I'm quite sure the 10% to 20% numbers are not realistic. Even with purposely abused motors/appliances.

The key is that the device is plugged directly across a dedicated breaker feed so it has no way to respond to any load variation. It can't know what the home's total reactive component is so it can't be dynamic in switching caps in/out based on system requirements which is what more sophisticated industrial units can do.

It has *some* effect in transient supression but is limited to a slight decrease in source impedance.

[Fnord]

I guess the only place we differ is that I can't even come up with a hypothetical with an uncomped motor where the device would make a material difference.

Considering that (1) I'm an electrical engineer, and (2) I evaluate electrical and electronic devices as part of my regular duties, I have to find ways to utilize seemingly useless devices, especially when some management wonk has bought 1000 or so units from a discount house that has since gone out of business. It's either find a way to make them useful, or pitch 'em into the Rohs recycler and write them off.

I think the product's claims are wildly exaggerated.

I think that's an understatement! :D

[marting]

I think that's an understatement! :D

You might be right. I'm a EE as well with a mix of designs as well, from nv amps to kv servos.

When I first read the claims I thought it must be some sort of adaptive device that measures the reactive and real load then comps out the reactive. I was wrong. When I downloaded the "supporting" docs and installation instructions it became clear that it had to be a fixed capacitor bank, probably producing 5 to 10 amps compensating reactive current, designed to offset typical home residual motor reactance. This because the thing was attached only across the line thus no method to measure demand let alone phase.

Another problem is that the wiring from the utility meter to the breaker panel is the only area where current, and hence ohmic losses, would be materially reduced. Those wires are pretty heavy gauge so we aren't talking anything significant.

There is an effect to the utility company since the line current from the utility is decreased by PF comp. That amount, while larger than energy savings in the home, is still fairly small for residences and hence isn't measured for tolling. Large industrial plants are PF checked as the impact on the utility becomes material so they impose surcharges on PFs that are too low incenting the plant to use comps. A plant can also achieve saving by placing the PF comps near the inductive load reducing the ohmic loss from there back to the meter. Comping near the meter doesn't save internally but does keep the utility happy and prevent the surcharges.

[bjb]

Considering that (1) I'm an electrical engineer, and (2) I evaluate electrical and electronic devices as part of my regular duties, I have to find ways to utilize seemingly useless devices, especially when some management wonk has bought 1000 or so units from a discount house that has since gone out of business.

OK, if you're going to dig for an application, how about this? Suppose you run your A/C day and night all summer long. The various compressor and fan motors are all single-phase induction motors and the line voltage at your house runs a bit low. Also, the wiring is old or maybe you have aluminum wiring in your house, so there is an extra few volts of voltage drop at large loads.

As I recall, the power factor for an induction motor is around .9 or so but it drops when the input voltage drops. Perhaps in this extreme case the caps can help bring up the power factor? I don't have a PSpice model for an induction motor so I can't try it out myself.

[Fnord]

PF is a factor of frequency, phase, and reactance, not amplitude -- there is no amplitude component in calculating PF from load impedance.

Otherwise, lowering the line voltage might affect the apparent load, which in turn could affect the PF.

If you can measure the PF, and install just enough capacitance to bring the PF back to 1, and then run those motors continuously with constant loading, the PF will remain at 1 all summer.

If you can have a device on your house mains that monitors the PF, and adjusts the capacitance accordingly, then the PF will remain constant no matter what the loading may be.

[illogical]

they did release these little things that screw into lamp sockets. they sorta work, albeit dimly. i assume they are diodes or triacs, and they "flicker".

if you want to save a lot of money, cancel your electric service.

[marting]

This article is laden with both error and lack of specifics in many critical areas:
http://www.power-save.com/Power-Save_study.pdf pg. 4

Due to the reduction in the total current, the power loss (I2 total x R1) in the resistance R1, between the wattmeter and the ABET-2201, which varies from house to house, is also reduced. This is the instantaneous power saving that is achieved by installing the ABET-2201. It is important to note that i) the resistance R1 will depend on the locations of the Energy-meter and ABET-2201, and ii) the power saving is proportional to the square of the reduction in the current brought about by the ABET-2201.

the power saving is proportional to the square of the reduction in the current...
Um, no it's not.

And this was put together by two people with doctorates? Yikes.

[CardZeus]

Many years ago you could buy (from dubious sources) a device which was basically a modified doorbell chime - connect it to your electricity meter and it would either slow down, stop or run backwards. One had to be careful not to overuse it as the prospect of the meter reader finding your usage was negative would be hard to explain. (Not that I ever used one...;)

[a_unique_person]

Many years ago you could buy (from dubious sources) a device which was basically a modified doorbell chime - connect it to your electricity meter and it would either slow down, stop or run backwards. One had to be careful not to overuse it as the prospect of the meter reader finding your usage was negative would be hard to explain. (Not that I ever used one...;)

An electrical engineering lecturer once told us about a simple circuit that would fool an electrical meter. I think the meters are more advanced these days.

[Fnord]

An electrical engineering lecturer once told us about a simple circuit that would fool an electrical meter. I think the meters are more advanced these days.

Oh, there is absolutely no way to fool an electrical meter ... no way at all ... none whatsoever ... there are no such secrets ... move along ...

;) (Tesla Coil)

[DRBUZZ0]

Oh, there is absolutely no way to fool an electrical meter ... no way at all ... none whatsoever ... there are no such secrets ... move along ...

;) (Tesla Coil)

Oh of course there is... you could just bridge it with some resistors to bypass the meter. I say resistors as opposed to just copper wire, because I think the electric company would be rather suspicious if the meter was reading zero consumption or something close to it. So you best not divert too much current...

Not that I'd condone that anyways...

[a_unique_person]

Oh of course there is... you could just bridge it with some resistors to bypass the meter. I say resistors as opposed to just copper wire, because I think the electric company would be rather suspicious if the meter was reading zero consumption or something close to it. So you best not divert too much current...

Not that I'd condone that anyways...

It was just a big capacitor or inductor to change the phase of the power signal, IIRC. That was enough to put the meter off, without looking too suspicious, such as a straight bypass of the meter. Resistors would not work as easily, also IIRC. I think he said some electrical devices inherently deceived the meters, due to their inductance.