Something to ponder

Add some series resistance between the source and the load. That series resistance causes a voltage drop of IR. An increased current reduces the voltage across the load. We should never assume a perfect source

Lets drop the voltage to a machine shop, the amperage will increase significantly, as the motors can't reach full speed, they run with starting amperage. Now we know where hurting our expensive motors and wasting energy, lets look into the bad advise they give us on our highbay flourescent tubes. They want us to decrease the line voltage thinking the amperage will follow the decrease, but we still have the electronic ballast and fixed wattage bulbs. This bad advice will increase voltage drop waste and actually increases line current, which in return induces even more voltage drop and even more line amperage. Now we now we are hurting our ballast. Take 120v filament, voltage and amperage proportional, the voltage decrease gives a slight inefficiency, but we gain longevity.

The electrician's comment is based on an assumption that a 1000W appliance will always consume 1000W. Therefore, if the voltage were reduced, the current would have to increase in order to still have 1000W. What actually happens will be a reduction in current and a reduction in the efficiency of the appliance.

Your 1000 watt example is flawed as those are not a filament bulb type. Metal halide/high pressure sodium/mercury vapor/halogen, all these bulbs rely on a gas arcing. Apply only 100 volts these bulbs and they may not be able to ignite, and eventually overheat and bust. Best not to flirt with these bulb types, they can spit a ball of flames or slowly emit cancerous uv rays.

For resistive loads like a light bulb the current will surly decrease, but modern electronic appliances have power supply units that operate at wide range of voltage, so if the device is operating drawing a certain amount of power from the mains and the voltage at the mains drops then the input current intake must increase to continue to supply the required power to the device.

The same for CFL light bulbs

If your light bulb is a simple incandescent job (though huge - 1kW is a one bar electric fire, not a light bulb) Ohm's Law applies. I=V/R. Since R is constant (ignoring thermal effects) if V is halved, I is halved. This assumes a simple resistive load, of course (a toaster, for instance). It's not so simple if the load's a discharge lamp or a switching power supply like a mobile phone charger.

One cannot ignore the filaments thermal effect light bulbs become more resistive as they heat up, reduce the voltage and the filament would cool down and loose resistivity and thus draw more current. The same applies to electric elements in stoves. The big clanger is anything with a motor, while the lights and elements will draw a little bit more current, dropping the voltage to a motor will definitely cause it to increase its current draw to the point where the voltage drops low enough it will stall, and draw enough current through it to burn it out.

@BillyT if the current actually increases then the temperature must surely increase due to I^2R losses and which intern increases the resistance of the filament.

@selman for lamps this is true, but it never quite makes it. For motors, basically the slower they turn, the less magnetic reluctance (read magnetic feedback) is generated to cancel the incoming current this lowers the inductive reactance of the motor thus the higher current persists, if the voltage goes low enough, the reactance is almost the DC resistance of the motor and that is when it can and in most cases will burn out. It is similar to having too large a load on an electric motor and having locked rotor current.

Yes I agree. Because when the rotor of an AC motor is running way below synchronous speed the equivalent rotor resistance as seen from the stator will decrease dramatically, drawing much more current.

Fast engine idle can use less gas than slow engine idle? Less time to draw fuel into cylinders on fast idle plus more startup horsepower when load on engine comes. As you guessed, some can only see slow engine idle as best.

@imccoig That depends on the camshaft. The pre 80's American auto were anti rpm, anti horsepower, and pro torque. They kept the exhaust valve closed longer, good for low end torque, bad for high end horse power.

I recon that this comment is based on empirical evidence of the sparks (electrician) on the job . its rational to a point when one considers that to a certain extent the supply is not an infinite current source. If this is taken into account the sparks logic works well. An significant current drain would lower the voltage where there is a finite current source. However, this depends on if the location is Europe or USA style supply. 110 volts is much lower resulting in the need for thicker wires. Its inherently less efficient than 240 volt supply systems. I would suppose that the EU system suffers less from volt drop caused by length of wire run in ring main installation. It would also be less responsive to the effect purported by the sparks. Im not up for trying out this as an experiment with 240 volt mains supply. But, one could use EC constant current generators to investigate!

@Zorgrian I can attest to the truth of the matter,this is also from living and working in a 240V environment.

@BillyT, in which way do you attest to the matter? Do you say that current drop could give rise to a volt drop, or be caused by the same. Or, do you believe this is not possible!

@Zorgrian, I attest that if the voltage across most lighting and or electric motors drops, the circuit current will rise.

Yeah that would be normal with a well designed system such as they have in UK. No chance of limited supply

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