Recently I had a discussion if overvoltage in mains lowers electricity bills (a bit) or not.
With incandescent lamps it does but it shortens their life as well - so no actual savings here.
With fluorescent lamps on magnetic ballasts - not sure - the same as incandescents?
Electronic ballasts with a voltage stabilizer - that's a question, because the drawn current is lower [I = P (constant) / U (higher)]. But meters - both classical and new electronic/digital ones should take the higher voltage into account so probably no savings here.

Then we discussed on if you can steal energy by shifting the power factor (cos φ) with capacitors on incandescent lamps or not. While meters (both classical and new/digital ones) don't charge for reactive power (just for active power), so the light bulb doesn't do anything with reactive power. That was my point but my opponent thought otherwise.
Also removed compensation capacitors from a fluorescent fitting are just an "annoyance" for power companies (they must compensate themselves) but the meter counts the same. Or not?

Well, I would think that undervoltage would lower it, Not overvoltage.
If the power meter only charges for reactive power, Then you could  change the power factor to change the power being read by the meter.

Is there any way (Or formula) to calculate the life when an incandescent lamp is under/overvolted?

If you run a filament lamp over its normal voltage it will draw more power not less a fifteen watt bulb draws 15 watts with 240 volts applied but nearer 16 watts with 250 volts applied and of course bulbs are resistive so there power factor is unity just what the leccy company likes. Flourescents lamps draw more power  at higher applied voltage but measurements are more difficult to interpret but basically things only draw the correct power at the correct volts

Incandescents :

Higher voltage leads to higher current, allthough the current is not proportional tho the voltage, it goes up to a lower extent. Power goes up. It is more than proportional to the voltage, but less than proportional to the voltage squared



Fluorescents :

Vary between power going up and staying the same

Magnetic ballast Fluorescent (i am thinking choke ballast here) : The lamp voltage stay about the same, as it is determined by the arc. Ballast voltage go up, respectively does the current. So lamp power is proportional to the current. By how much ? That depends a lot on the ballast

With low voltage lamps (PL-S 9W, T8 18W, ..) The ballast voltage is similar to the line voltage, that is, the voltage across the ballast rise about the same as line voltage. So does the current, and so the lamp power

With high voltage lamps (PL-C 26W, to less extent T8 36W..) The ballast voltage is less than the line voltage, it is close to sqrt(line^2 - lamp^2). So the voltage across the ballast is affected more significantly than the line voltage, and effect of overvoltage on lamp current is more than linear, so is the power

Ballast losses of Magnetic ballast consist of coil wire resistance losses - proportional to the current squared, eddy currents - proportional to the current, and hysteresis losses that vary for each ballast design, and are generally related to the magnetic flux density in the core. I expect that coil wire resistance losses are the most significant ones, so generally ballast losses are proportional to the current squared

Generally ballast losses are not too big compared to lamp power - Most ballasts do well over 0.8 efficiency and that includes relatively inefficient "mini" ballasts

Electronic ballasts vary : Ideally they are a constant current source, so the lamp power does not change at all. Less ideal electronic ballasts may vary the current to the lamp to some (probably quite small) extent, so the amp power does change a bit

Electronic ballast losses depend on the power delivered to the lamp, so i guess would change i n about the same way as the lamp power



Mercury and MH :

Same as Fluorescent



HPS :

Allmost same as Fluorescent - Higher line voltage leads to higher voltage across ballast so higher current. But then the arc voltage goes up as result - So the lamp power is more sensitive to overvoltage and goes up more significantly



LED :

Most LED drivers are switching constant current supplies, and the LED voltage is determined by the LEDs and the current, so the power stays about the same regardless of voltage. With less regulated designs like filaments it may vary though



Heating/Cooling appliances :

Heaters are fairly constant resistance, so the current is proportional to the voltage, and power is proportional to voltage squared. In theory the power would go up to the most significant extent of everything described above....

But there is the catch : Those appliances are controlled by a thermostat. The thermostat clicks them off when a certain temperature difference is reached in a volume of substance : Air in a room, Water in a container, and so on. This happens when a certain amount of Energy is delivered, which equals to = Heat Capacity of the substance (constant for each substance) * Volume (not changed) * deltaT (set by the thermostat). The higher power heater will take more power, but then the thermostat will click it off faster. The overall amount of energy used stays the same

As energy is transferred over the boundary (heat leaking out through insulation out of the hot water container to the ambient), eventually it will get to the point that the thermostat clicks on again. The deltaT of the thermostat (Toff - Ton) is quite small, much less than deltaT between the water and ambient air. That means, that the temperature difference between the water and ambient air is fairly constant, regardless of whether the thermostat just clicked on or off. As heat conduction is related to deltaT, it depends on the thermostat setting but not line voltage. The users use the hot water the same way regardless of line voltage either, so the amount of heat "lost" from the container that way stays unchanged

Then how fast the heat is lost (energy over time = power...) is fairly constant. When the heater is in use for long periods (controlled by thermostat), it all averages out on that input power = loss power. So in average the input power does not change

About the same is going on with cooling appliances. Losses of the compressor motor may depend on voltage, but they are insignificant compared to the "thermal" energy transfers in the system, and those are constant



Motors :

The outout power of electrical motor depends ultimately on the mechanical load connected to it. Mechanical loads can be of many types : In some the torque is dependent on speed, in some less. One thing is obvious : If the motor is spinning the stuff with the same speed, the torque will stay the same

Most AC motors used everywhere are Induction motors, which speed is small bit below mains frequency, divided by number of poles in the motor. So as long as mains frequency is the same, motor speed will stay about the same as well, then the torque same and the power same



Some AC motors use brushes, in them the speed of the motor is not bound to the mains frequency. How much will such motors accelerate ? It varies. For parallel excitation motors (coil provoding the flux is powered directly from the mains) :

EMF = Constant (specific to the motor) * Flux * Speed
Flux = Constant (specific to the motor) * Voltage
Voltage = EMF + Voltage drop on coil windings

If we assume the losses on the windings as insignificant, we get :

EMF = C1 * Flux * Speed
Flux = C2 * Voltage
Voltage = EMF

then :

Voltage = C1 * C2 * Voltage * Speed
Speed = 1/C1C2

So the speed stays sorta constant

In reality, it is not as nice and speed does depend on voltage, allthough to not big extent

Most motors are fairly efficient, so while motor losses are affected by the voltage, their part in the overall world is not big



Electronics :

Everything with switchmode power supply, the output voltage is same regardless of input voltage, therefore output current (determined by the load and the voltage supplied to it from the power supply) is unchanged, and output power is unchanged

Most switchmode power supplies havefairly constant efficiency, so the power input is unchanged. Maybe goes less a bit, as power supplies tend to work more efficiently with higher input voltage, but this is only noticable when comparing the efficiency of "100..240V input" on 100V vs 240V, i doubt it will be noticable on small differences like 230V vs 240V..



Everything with linear power supply (transformer + linear regulator like 7805), the output voltage of the transformer and the rectified DC after that are proportional to the input voltage. After the 7805 the voltage is 5V, and the current is determined by the load. So the current through the load, adn through the power supply, is unchanged regardless of input voltage. If we disregard the transformer losses, then the power used by device is :

P = 5V * Iload + (RectDC - 5V) * Iload = RectDC * Iload

And that is linearly proportional to the input voltage

The transformer losses consist of coil wire resistance losses - Those are unchanged as the current is unchanged, and core losses like from eddy currents and hysteresis - The eddy currents are proportional to the input voltage, hysteresis can vary (depend on specific transformer design) but are probably insignificant compared to everything else

So overall the power goes up about proportionally to the input voltage

Well, I would think that undervoltage would lower it, Not overvoltage.

It draws less energy but it gives off much less light. Overvoltage makes them efficient like "halogens" but with a much shorter life.

If the power meter only charges for reactive power, Then you could  change the power factor to change the power being read by the meter.

No, meters charge only for active power. Reactive power is ignored by them.

Flourescents lamps draw more power  at higher applied voltage but measurements are more difficult to interpret but basically things only draw the correct power at the correct volts

Yeah, practically all non-stabilised light sources power would go up with the voltage. The question is the lumen gain and life shortening.

HPS : Allmost same as Fluorescent - Higher line voltage leads to higher voltage across ballast so higher current. But then the arc voltage goes up as result - So the lamp power is more sensitive to overvoltage and goes up more significantly

Why more sensitive? Is it because of the HPS arc tube chemistry?


Nobody supported nor tore off my theory that you can't steal energy with incandescents and capacitors (thus shifted to reactive power) - see the OP. ...?

Fluorescent, Mercury and (atleasdt the basic types) MH have fairly constant arc voltage after the lamp warmed up. Higher current through the lamp would be only that - Higher current to the lamp. Same voltage * higher current = higher power

HPS have arc voltage that depends on the temperature, as it affects whether more or less amalgam is evaporated into the discharge. Higher current through the lamp makes it heat up more, and then the arc voltage rises. Higher voltage * higher current = more significantly higher power



Ways i can think of to get free or cheaper electricity through a meter, without tampering with the meter itself :

 - Draw so little power that the friction of the meter mechanics holds the plate in place. This might work, but probably at no more than single Watts. Actually i think much less. No usefull savings - At electricity costs here, powering 1 Watt load permanently for a year cost about 1.2 Euro. The inconvenience of trying not to put any other load on the line for long periods (and therefore finding more complicated alternatives) can cost much more..

 - Draw impulses of high current, so that the meter mechanics dont have the time to respond. This most likely won't work. Mechanical meters are mechanically quite fast. They do catch things like inrush currents and short circuits - The plate "jumps" to a new position that same moment. In a short circuit, the voltage across the meter output is allready so much below line voltage that it is useless. Try to make shortrer pulse, then you need even higher current draw to get some energy, then the voltage you get will be next to zero. Make the pulses too close together (to get overall some usefull amount of energy), and you'll either accelerate the meter so it does measure what you do, or blow a fuse..

 - Smart meters may be more vulnerable. They sample the voltage and current at set times (iirc something like 256/sec), then for each sample, energy = sampled voltage * sampled current * time interval, and total metered energy is sum of all those energy bits. In theory if you pull pulses of current between the times where the meter is sampling, your energy is free... But if they were smart enough to include an analog lowpass filter in the current sensing circuit, then it'll measure the current correctly and you'll pay correctly



To tweak lamp efficiency you dont have to go that far

Incandescents - Use 220V incandescents on 230V

12V Halogens - Add a turn or two with free wire over the existing toroid transformer

Fluorescents and HIDs - Use the wrong ballast (for small tweaks use 220V or 60Hz ballasts on 230V 50Hz), add some little bypass ballast in parallel (liek a PL choke), change feedback resistors in Electronic ballasts

LEDs - Use a different driver or change feedback resistors



About Fluorescents and wrong ballasts, here is a thing i recall reading :

Industrial fridge, with lighting intended originally for 80W T12 lamps (same form factor as 65W, and some 65W lamps were actually rated 65/80W). The 80W FL choke is identical to 80W Merc choke

They put up 58W T8's there..

They say the lamps dont last long but they do reach fair brightness, despite being T8's inside a fridge

Ways i can think of to get free or cheaper electricity through a meter, without tampering with the meter itself :

 - Draw so little power that the friction of the meter mechanics holds the plate in place. This might work, but probably at no more than single Watts. Actually i think much less. No useful savings - At electricity costs here, powering 1 Watt load permanently for a year cost about 1.2 Euro. The inconvenience of trying not to put any other load on the line for long periods (and therefore finding more complicated alternatives) can cost much more..

That is the only trick that technically works, but as you wrote, too much effort for no significant gain...
The electronic meters do have such limit imposed artificially (so they do not register any minor cross talk in the system - the customer protection laws just do not allow that) and the minimum threshold tends to be actually higher than with most of the mechanical meters, as the mechanical usually do not have the friction by far that large as the standard allows, while the minimum power is really imposed just by that friction. With electronic the minimum power is explicitly programmed so, even rather high coupling (high current wires going nearby,...) does not cause any counting against the customer.

- Draw impulses of high current, so that the meter mechanics dont have the time to respond. This most likely won't work. Mechanical meters are mechanically quite fast. They do catch things like inrush currents and short circuits - The plate "jumps" to a new position that same moment. In a short circuit, the voltage across the meter output is already so much below line voltage that it is useless. Try to make shorter pulse, then you need even higher current draw to get some energy, then the voltage you get will be next to zero. Make the pulses too close together (to get overall some useful amount of energy), and you'll either accelerate the meter so it does measure what you do, or blow a fuse..

That does not work (well unless you really saturate the current coil, but that would then trigger the breaker). The disc inertia indeed can not respond to such fast pulses (it accelerates slowly), but it can not stop immediately either (it decelerates even slower). If you would do more detailed calculations, you will find out, these effects exactly compensate each other. Bottom line, if we neglect the friction, it registers exactly the same amount of energy as if the same amount of the real energy would be distributed over longer period of time.

- Smart meters may be more vulnerable. They sample the voltage and current at set times (iirc something like 256/sec), then for each sample, energy = sampled voltage * sampled current * time interval, and total metered energy is sum of all those energy bits. In theory if you pull pulses of current between the times where the meter is sampling, your energy is free... But if they were smart enough to include an analog lowpass filter in the current sensing circuit, then it'll measure the current correctly and you'll pay correctly.

The meters indeed do sample the values, but the signal is low pass filtered (the purpose of the filter is just the aliasing you are describing), so just the first harmonic passes through and passes correctly. For the accurate energy measurement you do not need more - as the voltage is a sinewave, so the only component that correlates with it (so the only one transferring power) is again a sinewave of the same phase.
If you drive such current, the in-phase sinewave component is zero, the meter will indeed register no power, but with a sinewave voltage you will even transfer no power at all (with the disturbance so strong it distorts the mains voltage the real power will even go backwards - the power of the higher frequency components goes from the load back to the mains).
The low pass filters are in both the current, as well as the voltage sensing circuit. In the voltage path the filtering itself is not necessary, but the filter function there is to match the phase shift of the filter in the current signal path, so when the voltage is in some phase towards the current, the signals on the ADC inputs are in the same phase relations as well.
The filters are only partially in HW, the biggest part is in the digital/SW, as the ADC's are made as sigma-delta type (with physical sample rate in 100's kHz range), where the decimation filter does the biggest part of this filtering job. The decimation filters reduce the sample rate from the 100's kHz to the 100Hz or so, while at the same time boosts the resolution from a single bit samples to at least 10 bit values for the power calculations, the result is then averaged out (well, the energy integration does the averaging) to get further resolution required for the correct energy measurement.

Another idea for analog meters is to draw so much power that eventually the needles for counting the power rotate enough to reset themselves(Like a clock going from 12 to 1).

Another idea for analog meters is to draw so much power that eventually the needles for counting the power rotate enough to reset themselves(Like a clock going from 12 to 1).

Well, here the meters do not use the needles to display the registered value, but numerical drums. And there are just too many of them - most common are 6 digits for kWh. You can never overflow that in only the meter replacements period (that is maximum 3 years, if I remember well; assume it is used for billing, so the legal mandatory calibration intervals apply for that).

Whateve reading you end up with youl get charged for it and also Medved is correct in saying that the highest 2 registers hardly ever move in the UK they are usually 1000 and 10000 KWh that takes a lot of effort to use that much