# Using inductors to reduce electric bill?

1. Mar 30, 2005

### mathrocks

Using inductors to reduce electric bill??

My professor today mentioned that if you place an inductor in one of the power outlets in your house will allow actually produce no power and will cause your electric bill to reduce since that specific outlet is causing the power meter to reverse its direction?? Sounds pretty dumb to me, but he was very serious about it. My question is will this actually work and if so, how can you still use that outlet if there is an inductor placed in it? It seems pointless to me since that specific outlet isn't doing anything in the first place and adding an inductor just does nothing again.

I tried searching google but nothing really showed up about using this technique, so I thought I ask the experts here.

2. Mar 30, 2005

### NoTime

A perfect inductor would draw no power, so in theory a perfect meter could run backward.

In practice:
1) There are no perfect inductors, so there will be a cost.
2) Some meters (maybe all nowadays) charge for reactive power, particularly commercial ones.

In the later case capacitors are often added for power factor correction.

3. Mar 30, 2005

### Averagesupernova

I'd say he is full of sh!t. Many household loads are inductive. Meters measure power and the direction it is flowing. If you want to make your meter run backwards you need to pull it and install it up-side-down or actually generate electricity and 'power the grid' so to speak. However, generating your own electricity in order to take money off of your bill is equivalent to driving your car in reverse in order to take miles off of it.

4. Mar 30, 2005

### faust9

You can do it. Shif the power factor with a capacitor(Large i.e. 100's of dollars capacitor).

5. Mar 31, 2005

### egsmith

The power factor? I am not sure how this would change things. Could you be more specific.

I did find this
http://en.wikipedia.org/wiki/Power_meter

This is my first time researching the implementation of a power meter but the description of the implementation make sense to me. The short non technical summary is, as current is passed through a wire a magnetic field is generated. The power company's meter takes advantage of this effect by placing some mechanical parts in the magnetic field generated by the electric current which is delivered to the house. The mechanical part gets a small nudge by the magnetic field causing a small movement. The sum of the nudges, or the distance the mechanical parts moved, or the power consumed during a period of time, is viewed (perhaps indirectly) by the power company and is in turn used to determine how to bill.

So I suppose if current was moving in the other direction (out of the house) the nudge would be in the other direction taking something off your bill (assuming there is no mechanism to prevent movement in the opposite direction).

Now here is why I don't think an inductor or capacitor would work to save money. You have to pay for the power that initially charges these devices regardless of their size. If one discovers a circuit that gives back some of this stored power (by temporarily reversing the current through an oscillation), so be it. But that does not cancel the power used by the rest of the house so the overall bill would be no lower. In reality it would actually be higher because the inductor and capacitor are lossy power storage devices.

6. Mar 31, 2005

### chroot

Staff Emeritus
Power companies typically charge residential customers without regard to the power factor (the meter is not sophisticated enough to consider the power factor). If you shift your power factor so your load is more reactive and less resistive, your meter will record less current and you will be charged a smaller sum. It's correct. If your power factor is -really- wonky, though, the power company might eventually notice. Reactive loads still cause currents in their transmission lines, and they still lose energy in the heating of those lines.

Large industrial centers have large inductive loads (i.e. motors), and the power company -does- charge them in consideration of their power factor. As a result, these businesses add capacitor farms to their mains to put their power factor back close to one.

- Warren

7. Mar 31, 2005

### SGT

Power meters measure only resistive power. If you have a reactive load (inductive or capacitive), the power drawn by this load will be returned to the electric network, so the total reactive power in your home is zero and not negative.
The problem is that the electric cables between the power company and your home will be heated by the extra current drawn by your reactive load, so power is lost by the company and nobody benefits from this. This is why power companies charge large customers with reactive loads.

8. Apr 5, 2005

### Antiphon

Briefly, power flows back and forth in and out of your house if the power
factor is not 1. You typically only pay for the part of the power that flows in.
But because of inefficincy in the wires, you end up paying for a small part of
the power flowing out of the house that's lost to the inefficiency (resistance)
of the house wiring. By adding a capacitor to balence an inductive house load
you can convert the power to inflow-only and eliminate the losses.

9. Apr 10, 2005

### MarkT

Regardless whether the meter is inserted correctly or not by adding inductors there will be some true power this would be added to the existing load and would add up the cost of your bill.

Years ago a meter could be inserted upside down and would run backwards however most power companies today use much more advanced meters that then and they will NOT run backwards.

Remember this, most states have meter tampering laws and the power company gets upset when some one cuts a seal and removes the meter.

On residential services usually there is one transfromer for 3-5 sometimes more homes and the power company knows what the loads are , this is one of the reasons if you upgrade a service, of significance, you are suppose to let the PoCo know the standard transformer may have to be upgraded too.

If you start adding inductors at outlets and plug in reisistive loads the total impedance will shift to be more and more resistive, so aside from the aggravation of adding inductors where is the cost savings.

10. Apr 10, 2005

### chroot

Staff Emeritus
This is false; you would be changing your power factor from its normal state, nearly all resistive, to being much more reactive. You would have less real (metered) current flowing.

- Warren

11. Apr 10, 2005

### MarkT

from: Warren;This is false; you would be changing your power factor from its normal state, nearly all resistive, to being much more reactive. You would have less real (metered) current flowing.

We are talking residential, the question was to save money on his electric bill. Most residences have a Fp of .95 up to 1 where are you going to save the money.

Oh and we are not going to add resistive circuits ? What are you going to use to see with ?... Candles ?

Where would the savings be, say on a 200 amp 1-phase service 48kva, thats an awfull lot of inductors. What size inductor would you need for 1-1800 va receptacle thats 80 % of a standard 20 amp breaker and you got a receptacle every 12 linear feet.

Even if it were possible you dont think with 5 other houses on the same transformer that there wouldn't be a heating problem at the transformer, that reactive power has to go somewhere.

Last where do you find perfect inductors ..

Last edited: Apr 10, 2005
12. Apr 10, 2005

### chroot

Staff Emeritus
We're not talking about unplugging resistive devices either. Leave all your stuff plugged in, but then plug in an inductor in an unused outlet. The end result is that your meter will not run as fast, and your electric bill will be smaller.
You're discussing practicality. We're discussing theory. The reason no one actually does this is because, as you point out, there are practical concerns that make it unattractive. That doesn't mean, in theory, that it wouldn't work.

- Warren

13. Apr 10, 2005

### Cliff_J

Warren, how would the meter run slower and not faster? An additional inductor in parallel with existing ones would reduce overall inductance and thus improve the power factor seen by the meter even if there is no improvement at the loads. Plus you have an additional load flowing current, and our meter is more a current meter than a power meter anyways, double negative.

14. Apr 11, 2005

### MarkT

Given:
25kw load at unity Fp. @240 volts Line-line total current is 104.2 Amps.

Using the Inductance theory we will add 25Kvar of inductance to the house.
this comes out to be 37.36Kva@45degrees.
I = 147.3 amps @ 45 degrees.
the components are 104.2 +j104.2 amps
the true power has remained the same.

by adding the inductance you have NOT changed the true power but have managed to overload the transformer because it is now supplying 37Kva instead of 25Kva.

this goes back to the law of conservation of energy, 'You don't get something for nothing'

Theory has its place. Back in the 70s when the gas crisis hit , there were rumors of somebody had invented a Carboretor that would give you 50-60 miles per gallon, but it wasn't available because the big oil companies and the Auto makers bought it in order to sell their products. This went by the supposition that a 6500 pound Cadilac would get the same gas miliage as a Pinto barely 2000 pounds. Bottom line there is only so much energy in a gallon of gasoline no more no less, the theory defied logic.

So unless you remove the resistive loads, true power, and somehow replace them with all Reactive loads, the true power will not change. And whether you can fool the meter with inductive loads I am not sure, because the Utility knows exactly what their Fps are everywhere.

Mark

15. Apr 11, 2005

### Averagesupernova

Cliff, the meter certainly does measure true power and not current. Reducing total inductance by adding inductors in parallel does not improve power factor. ALL inductors have a resistive component and by paralleling more of them in you are adding this resistive component as well. The only way to improve the power factor is to add the opposite reactance to the system. Usually capacitors are added because the norm is having inductive loads. A residential customer does not pay for apparent power because the meter is unable to measure it so there is no reason to attempt to correct for the power factor in the typical home. However, you WILL reduce the total current through the transformer and meter if you keep your power factor at 1 but you won't save anything on your bill. For those of you unable to grasp this you need to have a look at basic AC theory.

MarkT, you mentioned in one of your posts that "that reactive power has to go somewhere". It really doesn't go anywhere. The transformer will heat up because of the IR loss, but it's not like the transformer is actually absorbing all of the apparent power as if it were a load. The generator doesn't absorb it either. I suspect you know this, I just wanted to clarify it for the others.

16. Apr 12, 2005

### Cliff_J

Averagesupernova - if we're adding an RL element that was able to alter the phase angle seen by the meter it would alter the power factor as seen by it. I could easily be incorrect on that one by confusing the vectors on series versus parallel RL though and if our PF is .95 then this is a very small shift to influence anyways.

Regardless of that, the new inductor would be an additional load on the system absorbing power, don't see how the meter will record anything but the true power absorbtion of the new load since in my mind I picture the meter as the voltage source and the rest of the house as a seperate parallel leg in the circuit. Additional power means the meter runs faster and the bill is higher.

17. Apr 12, 2005

### Averagesupernova

Cliff. I didn't say that adding inductive loads would not alter the power factor. I said it would not IMPROVE the power factor. I should clarify that by stating that my assumption was that the complex impedance was already inductive and not capacitive.

Yes I know the new inductor is a load and will absorb power. But not the full amperage it is drawing multiplied by the voltage across it. You are right, the meter will measure TRUE power dissipated in the inductor.

There seem to be some people posting here that think since you move the power factor away from 1 then the meter will change. The meters DO NOT work this way. Yes, the more inductors you add in parallel the more inductive current that will flow through the meter and the ratio of that to resistive current will become higher. But that does not mean that the resistive current, which is a DIRECTLY related to true power will change. It still takes 100 watts of true power to light a 100 watt light bulb and you WILL pay for it no matter what other sort of load you attempt to put on it.

Also, the power lost in the heating of the inductors because they are not 100% efficient will be paid for.

18. Apr 13, 2005

### MarkT

This is the thing with sizing motors. because reactive power the current is still higher than it would be with an Fp of 1. So the conductors must be sized bigger in order to accomodate this.

If the conductors are size just for the nameplate of the motor , there will/can be an excessive voltage drop on the lines, and especially at start up.

19. Jun 5, 2005

### Xodar

Let's start from the beginning...

Domestic loads are primarily resistive, with some inductive component. Your domestic power meter measures Watts - "resistive" power if you like, "average" power if you want to be consistent with electrical engineering.

Industrial loads are more significantly inductive and they are metered with both Watt Meters and VAr Meters to measure the reactive power. Because they are charged for both, but only benefit from the Watts, they install capacitor banks to balance their power factor. How does this work? The VArs in the inductive load are stored in the electric field of the windings on every half cycle and pushed back to the electricity authority on the other half. By putting a capacitor bank in, the energy coming out of the coils (magnetic field) is stored in the capacitor bank (electric field) and then retrieved in the next half cycle. This means, after the plant is initially started up, there is very little flow of reactive power through the VAr meter and the costs are reduced.

How does this affect the domestic situation. IF you were charged separately for reactive and average power, then you could finetune your domestic load with CAPACITORS to achieve the same result. Unfortunately, your electricity authority only measures the average power at your house and uses a factor in the charging structure to allow for the typical reactive load in a house. So even if you negated all of your inductive loads you would still be paying for them.

Theoretically possible to balance out your power factor? Yes. Actually achievable? Yes. Reducing cost for a domestic situation? No.