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?


9 Answers 9


SCAM WARNING. These "energy saver" devices usually are simply capacitors, and they don't save you any money. Typical customer energy (usually billed in kilowatt-hours, 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 energy 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 a meter: one with two dials. One dial is used to bill the customer for real energy consumed, while the other is used to bill wasted or 'reactive' energy. 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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    \$\begingroup\$ 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/… \$\endgroup\$ Oct 20, 2015 at 20:22
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    \$\begingroup\$ 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. \$\endgroup\$
    – Transistor
    Jan 21, 2016 at 20:45
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    \$\begingroup\$ @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. \$\endgroup\$
    – wbeaty
    Jan 22, 2016 at 6:37
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    \$\begingroup\$ This is true for old-fashioned rotating disk meters, but it's worth noting that it may or may not be true for the modern digital meters. I've found nothing conclusive on the subject. \$\endgroup\$
    – marcelm
    Jul 16, 2017 at 9:49
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    \$\begingroup\$ @marcelm There's nothing "conclusive" because the electronic meters have the full picture, and how that translates into the count on the display as well as into the billing is a matter of configuration settings. Smart energy meters measure the complex power, i.e. both the active and reactive components. A power company can choose not to bill residential customers for the reactive power. Typically, that power is a constant fraction of active power, and they compensate for it with capacitor banks at the substation. \$\endgroup\$ Jun 29, 2022 at 13:44

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.


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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    \$\begingroup\$ @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. \$\endgroup\$
    – AndrejaKo
    Sep 5, 2011 at 20:00
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    \$\begingroup\$ 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. \$\endgroup\$
    – AndrejaKo
    Sep 5, 2011 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.


Adding series reactance, resistance or anything in series after the meter will only cause a power dissipation. Your device will still draw the power it needs. Lowering the input voltage will cause the power supply to draw more current. Lower voltage lets say in a toaster oven would cause lower heating ability and need to cook something longer. It all is the same in the end except the extra power you now draw to satisfy the losses created by the series reactor. The only way you can decrease the amount of power you use is to decrease the series resistance. A home wired with all 12ga wire as compared to a home wired with 14ga would be more efficient. Would the cost of using only 12ga wire offset the power savings, maybe, maybe not. The best way to save money on you power bill is solar, but the ROI is still around 15~20 years. But with the development of fusion reactors, all of this will be moot in the next ten years as power will no longer be an issue. The only issue then will be the delivery grid.


Evergreen topic, so why not?

Short answer, no. Short on time? Read here: http://www.nlcpr.com/Deceptions1.php

Why? Household utility meters measure real power (kW) only. They're not influenced by reactive power. This includes both mechanical 'spinning disk' meters and the newer smart meters. (Mathematically, they take the instantaneous product of I*V, so if I and V are not in phase, as would be the case with a reactive load, the reactive I-V component cancels itself out over time.)


  • real power, expressed as kW. Meter measures this and this is what you're billed for.
  • reactive power, expressed as kVAR. You're not billed for this.
  • apparent power, expressed as kVA, is the vector sum of kVAR and kW. This is what your feed needs to be
  • power factor = (real power) / (apparent power) = kW/kVA


  • With a non-reactive load, power factor will be 100%.
  • With a purely reactive load, power factor will be zero (that is, no real power is consumed.)

These 'energy saving' devices, like many scams, rely on a grain of truth. In this case, it's the adding of capacitance across an inductive load (like a motor), in the right amount, can improve power factor factor and thus reduce reactive currents circulating back and forth to the utility.

In fact, large industrial users employ banks of capacitors which they can switch in and out as needed to manage their power factor, so they can avoid a low-PF surcharge from the utility. Typically they try for PF greater than 80%.

Stated another way, using the right power factor correction reduces your apparent power (kVA), while your real power (kW - what the meter measures) isn't affected.

But these household 'energy saver' things aren't good power factor correctors. They only offer a fixed capacitance value, so almost certain it's not properly matched to the load. And besides, since no reputable companies offer them, they're often shoddily built and lacking appropriate safety approvals.

Can power factor correction (PFC) be be beneficial to a home user? From a ratepaying standpoint, not at present. Utilities don't monitor residential PF. But this could change. Since the time of this original question, regulations have come into play that encourage (and sometimes mandate) that appliances use power factor correction within the device itself. This is a big deal with IT gear for example. And, smart meters can calculate PF and report it to the utility.

If you're building an off-grid, battery-inverter or generator backup system, you do need to care about power factor: you need to provision your supply based on kVA (apparent) power, not just kW (real) power. If your overall PF is good (above 80%, say) you can provision your supplies to be smaller. To help achieve that you'd opt for appliances that have PFC built-in.

Related: Scam? Does having this capacitor wired as shown provide any surge protection?


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

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    \$\begingroup\$ 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. \$\endgroup\$
    – user3624
    Sep 2, 2013 at 14:34

In fact a capacitor in series with a heater can lower the consumption of the heater and create a simple heat regulator. My 1000 watt ventilated heater is now set to 450 watt with this method.

  • \$\begingroup\$ This is true, but the "power saving" devices are not installed in series. \$\endgroup\$ Jun 29, 2022 at 13:45

You can do things to reduce electrical bill.

If you connect a well dimensioned capacitor in series with your TV, PC, TV box, lamps, DVD, phone charger, notebook charger, etc., you will see that the device will work fine with less power. My LCD TV is working fine at 180 VAC. I am in Europe and the network voltage is 230 VAC but with the series reactance you can lower the voltage applied to your device.

  • \$\begingroup\$ You won't lower the voltage, you'll shift the phase between the current and the voltage. \$\endgroup\$
    – Voltage Spike
    Aug 31, 2017 at 2:06
  • \$\begingroup\$ If you lower the voltage to many electronic devices, they will increase the current drawn to have the same power. Your TV is probably drawing more current and actually increasing your electric bill! \$\endgroup\$
    – DoxyLover
    Aug 31, 2017 at 6:31
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    \$\begingroup\$ Electronic loads are essentially constant-power, and if you lower the voltage, the current goes up. \$\endgroup\$ Jun 29, 2022 at 13:47

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