# If the energy meter sees a reactive load, will it register a lower reading?

At an exhibition on home appliances, I saw this guy hawking his 'energy saver device'. The guy claimed that merely plugging his instrument into the supply would reduce the consumption reported in an energy usage meter by a fourth. He mentioned something about the load seen by the meter being non-resistive ...

Not sure if this is the right forum to post the question (feel free to vote for the question to be closed)... Anyhow I find myself wondering - Is this just a sales-pitch (no warranty/guarantee on the device) or is there some fire to the smoke?

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SCAM WARNING. These "energy saver" devices usually are simply capacitors, and they don't save you any money. Typical KWH meters are not affected by adding a capacitor.

The scam works like so:

1. Many loads in your home are inductors (fridge motor, furnace fan)
2. If you install a capacitor which has just the right value, the current in the power grid leading to your home is reduced.
3. Con artists correctly claim that some energy somewhere is being saved.
4. Con artists correctly claim that industry uses these capacitors to save money
5. Con artist sneakily insinuates that this somehow saves YOU money
6. As evidence, con artist supplies testimonials rather than basic lab test results.

So why doesn't this save you money? It's because, while motors do draw extra unnecessary current, the KWH meter on the side of your home is designed to ignore that extra current! Adding a capacitor doesn't change your electric bill.

So energy is really being saved, right? Yes: it's energy which otherwise would heat all the power lines between the company generators and your home. The extra capacitor doesn't cause your motors to use less energy. Instead it relieves some load-current on the power grid. The electric company benefits from this ...but the homeowner doesn't!

Why then do factories use these Power-Factor Correction capacitors? Ah, for most huge industrial customers, electric utility companies install a different type of KWH meter: one with two dials. One dial is used to bill the customer for real KWH consumed, while the other is used to bill wasted or 'reactive' KWH. These industrial meters do detect the excess current drawn by induction motors. The industrial customers are charged for the unnecessary heating of the power grid. If they install just the right value of capacitor, they can reduce their electric bills.

And this brings up one last bit of info. To reduce the excess current in the power grid, the capacitor has to be just the right value!

If you have no induction motors in your home, then a PFC capacitor is less than worthless. Adding a PFC capacitor will INCREASE the wasted reactive current, not reduce it. So basically that's part of the dishonesty: selling capacitors of an unknown value in order to cancel out the effects of an unknown number of induction motors ...which aren't being billed by the electric company in the first place.

Finally, what about #6 above? The testimonials? I suspect that these are genuine. If you were to install a very expensive PFC capacitor in your home, you'd be bringing in the "stone soup effect." You'd become very aware of any wasted energy. You'd start "helping" the device: turning off lights, turning down the furnace and the air conditioning, perhaps buying better windows and installing improved insulation. The expensive and worthless "stone" has turned into "soup." But you'd save lots more money if you skipped the PFC capacitor scam and just started turning down the hot water heater in the first place.

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As an addendum to this, power factor correction is done for household consumers... by legislating active PFC requirements for devices, as done in the EU: electronics.stackexchange.com/questions/47026/… – Fizz Oct 20 '15 at 20:22
" the KWH meter on the side of your home is designed to ignore that extra current"... can you cite your source? – jpcgt Jan 21 at 20:12
The kWh meter (not KWH which means kelvin-watt-henries) meter is as the title suggests a kWh meter. It does not measure reactive power which is measured in kVArh. Look up how power meters work. – transistor Jan 21 at 20:45
@jpcgt my source: the meter itself, taken apart. The mechanical meter multiplies together the instantaneous voltage and current by using coils to produce a 2-phase rotating magnetic field, using this to rotate an aluminum disk against drag from a permanent magnet. If the V and I waveforms are at 90deg (lead or lag,) then the two b-fields will be at zero phase, and the torque upon the aluminum disk is pure AC with zero average, as it would be for ideal capacitive or inductive loads. The disk just vibrates without turning, and the "imaginary" current is ignored. – wbeaty Jan 22 at 6:37

I think this is a perfectly legitimate question. Energy meters like on the side of your house or on the utility pole already measure real energy delivered. If you were to have a purely reactive load, like a capacitor for example, then the energy meter would not increase but you also wouldn't be getting any energy. The capacitor won't get warm.

But if you put a current meter in series with the capacitor, you'd see real current. How can this be? No laws of physics have been violated because the voltage and the current are 90 degrees out of phase, something you can't tell just be looking at either in isolation.

The utility's electric meter measures the integral of the voltage times the current, so measures only real energy delivered. The utility doesn't like highly reactive loads because it causes currents in the transmission system, which waste power by I**2 * R losses which they can't bill for. This is why large electric customers get charged in part by the power factor. This is basically a measure of how far off you'd be assuming the product of independently measured voltage and current gave you real delivered power. For a purely resistive load, the voltage and current are in phase and the power factor is 1. That's what the utilities like. The worst case are purely capacitive or purely inductive loads. The voltage and current are 90 degrees out of phase, so no real power is delivered, and the power factor is 0.

In general, the power grid looks somewhat inductive. Utilities combat this various ways. These include banks of capacitors, nagging legistlators to force appliances to have better power factors, and running their generators a little out of phase. The utility term for the latter is "reactive power". In most cases you're actually doing the utility a favor by plugging a little capacitance into your outlet.

Unfortunately there are lots of snake oil salesman ready to exploit the fact that most people don't understand that independently measured voltage times current doesn't give you delivered power. There may be some slight saving to you in presenting a better power factor, but 25% "less electricity used" sounds like pure BS, and your electric meter already measures real delivered power anyway.

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It depends on the meter. If the meter reads only active power (P), then plugging in a non-resistive load wouldn't show up on the meter.

If however the meter reads the total power (S which is $\sqrt{P^2+Q^2}$, where Q is the reactive power), then if your house consumes a lot "inductive" power, plugging in something with "capacitive" power would indeed lower the total reactive power ("inductive" power cancels with "capacitive" power). If in the equation above you minimize Q, you would minimize S, which would minimize your bill.

I don't know how they bill electricity over in the US, but if the meter reads active power only, then this device would be a sham. If it reads S, which I highly doubt, you MAY experience a minor bill.

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Cool. Thank you. For the record (+: I'm in India – Everyone Sep 5 '11 at 19:29
@Everyone If you really want the answer to the question, you'd need to check this with someone who is familiar with the internal operation of the meter or to check the energy bill. In addition to meters which read only active power or only apparent power, there are meter types which read both active and reactive energy and which display the values. If that's your case, then it would be easy to see how much the device will help. – AndrejaKo Sep 5 '11 at 20:00
Also do note that the reactive power may be more or less expensive than active power and that there might be incentives in place to minimize it. For example I get considerable increase in reactive power cost if it's over 5% of total consumed power. If you have a power meter that measures only the apparent power, then you'd have to determine how much power is active and how much is reactive and there's no simple and easy way to do that for the whole household. If you have a meter that only measures the active power, then the device won't ever pay for itself. – AndrejaKo Sep 5 '11 at 20:05

One point i want to add is that , if you somehow calculate the right amount of capacitor bank to be used to cancel out the excess reactive power in domestic grid, then you can save some money in using thin wiring than the thick wiring that you would have used when you install some large appliance in your home. the reactive current may not register itself in the domestic meters but it will flow in the household wiring.

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Any device installed after the utilities meter can't and never can reduce energy consumption! Simple as that!

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A switch can. Turn off the switch, and energy consumption goes to zero. I could come up with lots of other examples, but if they are useful is debatable. "Never" makes for such a blanket statement. – user3624 Sep 2 '13 at 14:34