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Can someone clear up a few points for me please. With voltage drop when the voltage is decreased by a % will the current increase to make up for the loss in voltage to give us the same output?

And the more current we draw the greater the volt drop?

Thank you

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Hi - as a starting point we’ve Ohm’s Law which says the current through a conductor is proportional to the voltage across it. If the V goes up, the I goes up. And if the V goes down, the I goes down. So for resistive heating elements like a kettle the heating power In Watts goes down if there is voltage drop on the supply line.

as he said ^^^^^, and yes, the volt drop invreases proportionately with increase in current. as it also does with length. mV/m x A x L. ( the longer the cable, the higer the resistance. ).

As above, the only way the current can increase to maintain power is with the use of electronics, like what is used in power supplies, inverters and the like. So examples of devices which will maintain its power levels are: TVs, modern a/c units, modern heat pumps, motor control using VFDs (inverters), LED drivers, etc etc

Different loads behave in different ways. As P&S points out the current is most likely to increase if there is electronic regulation of the load involved, which will try to maintain constant power output. If it works at constant efficiency, that means constant power input hence increased current.

True resistive loads e.g. heaters; current decreases in proportion to voltage, therefore power decreases with the square of the voltage. However, heaters with thermostatic control will maintain the same long-term average power by staying on for longer each cycle of the thermostat.

Tungsten lamp loads; current decreases with voltage but not in proportion, because of the positive temperature coefficient of resistance of the filament. As it cools, its resistance falls, which partially compensates for the reduced voltage, tending to maintain the current more nearly constant. The light output falls drastically though, as this is strongly dependent on temperature.

Switched-mode power supplies (most modern electronics inc. LED drivers); Active regulation maintains constant power, therefore current rises in inverse proportion to the voltage.

Motors; current and power factor change in different ways depending on the type of motor and load. A vacuum cleaner (universal motor driving a fan) for example will slow down and the current will fall. An induction motor driving a conveyor will maintain near constant speed and the current and power factor will both increase.

Voltage drop along a conductor (VXI=P) so if said voltage was reduced due to voltage drop. The new voltage is proportional to current so as the voltage has been reduced so will the current causing the overall power to also be reduced?

Voltage drop along a conductor (VXI=P) so if said voltage was reduced due to voltage drop. The new voltage is proportional to current so as the voltage has been reduced so will the current causing the overall power to also be reduced?

say you start with a 240V supplty and a load of 100 ohms. I = V/R = 240/100 = 2.4A

if you dropped 10% , 24V due to cable length (resistance) you then have 216V at the load. the load is fixed at 100 ohms, so again I = V/R = 216/100 = 2.16A.... so the kettle will take longer to boil.time enough to read a chapter i BS7671.

But for the kettle example for the resistive loaf the current has gone up so the power is kept as a constant so as voltage drops current increases due to loss in voltage.

So 216V x 2.16Amps= 466.56Watts
250volts x 2.4Amps = 576 watts

So the voltage drop ultimately leads to Power losses in the Kettle and more current to be drawn in the conductor.

No, that is the wrong way round. For the kettle example which is a basic resistor then the total loss will always be proportionally split between it and the cable, but as the supply voltage increases both increase.

For example, if your cable is 1 ohm and kettle is 22 ohms then total is 23 ohms and at 230V you have 230V / (1R + 22R) = 10A.
Cable loss = I^2.R = 10 x 10 x 1 = 100W
Kettle power = I^2.R = 2200W
Ratio = 2200 / 100 = 22 (as for original resistance)

Repeat for 250V and I = 250/23 = 10.67A
Cable loss = 10.67 x 10.67 x 1 = 113.8W
Kettle power = 10.67 x 10.67 x 22 = 2505W
Ratio = 2505 / 113.8 = 22 (to rounding)

Again it is obvious it will always be that as (I^2.Rcable) / (I^2.Rload) = Rcable/Rload

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For a restive load it is straightforward to compute the other case, when you have a fixed supply and add cable resistance because you have:
I = V / (Rcable + Rload)
For other loads as Lucien Nunes has pointed out there is no fixed 'Rload' value but it depends on the supply voltage.

In that case the old method of solving this would be to plot the load-line (i.e. current versus voltage) of the load, and the the slope of the supply (voltage against current), and where they intersect is your operating point (i.e. combination of current and voltage that simultaneously satisfies both the characteristic of the supply impedance and that of the attached device).

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Also important aspect is for a fixed cable resistance (i.e. typical case from your design) the power loss goes up with the square of the load current which can become expensive over a long period.

Since your loss is small compared to the load (by design it has to be less than 5% or 3% volt drop for lights) you can approximate that as the loss is proportional to the square of the planned load (as the impact of the volt drop on the load draw is only a few percent whether pure resistive or "constant power" electronic, etc).

No, that is the wrong way round. For the kettle example which is a basic resistor then the total loss will always be proportionally split between it and the cable, but as the supply voltage increases both increase.

For example, if your cable is 1 ohm and kettle is 22 ohms then total is 23 ohms and at 230V you have 230V / (1R + 22R) = 10A.
Cable loss = I^2.R = 10 x 10 x 1 = 100W
Kettle power = I^2.R = 2200W
Ratio = 2200 / 100 = 22 (as for original resistance)

Repeat for 250V and I = 250/23 = 10.67A
Cable loss = 10.67 x 10.67 x 1 = 113.8W
Kettle power = 10.67 x 10.67 x 22 = 2505W
Ratio = 2505 / 113.8 = 22 (to rounding)

Again it is obvious it will always be that as (I^2.Rcable) / (I^2.Rload) = Rcable/Rload

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For a restive load it is straightforward to compute the other case, when you have a fixed supply and add cable resistance because you have:
I = V / (Rcable + Rload)
For other loads as Lucien Nunes has pointed out there is no fixed 'Rload' value but it depends on the supply voltage.

In that case the old method of solving this would be to plot the load-line (i.e. current versus voltage) of the load, and the the slope of the supply (voltage against current), and where they intersect is your operating point (i.e. combination of current and voltage that simultaneously satisfies both the characteristic of the supply impedance and that of the attached device).

So a 2200w kettle will have have more power at a higher voltage? As per example they are both 22 ohms resistance so identical. So in a practical example if we buy a kettle which states that as it’s load and we use it on a 230v supply it will not be as powerful as it would using a 250v supply?

So a 2200w kettle will have have more power at a higher voltage? As per example they are both 22 ohms resistance so identical. So in a practical example if we buy a kettle which states that as it’s load and we use it on a 230v supply it will not be as powerful as it would using a 250v supply?

Can someone clear up a few points for me please. With voltage drop when the voltage is decreased by a % will the current increase to make up for the loss in voltage to give us the same output?

And the more current we draw the greater the volt drop?

The current is always the same no matter where you measure it. It is determined by the overall resistance of the circuit from end to end. For an individual device in the circuit the voltage drop depends on the resistance of that single device (Ohms Law : V = IR) and the sum of all the individual voltage drops in the circuit is equal to the total voltage accross the circuit from end to end. That's Kirschoff's Voltage Law.

Also important aspect is for a fixed cable resistance (i.e. typical case from your design) the power loss goes up with the square of the load current which can become expensive over a long period.

Going off on a tangent, this is why the DNO typically likes to keep the supply voltage as high as they can. A small change in voltage can cause a disproportionate change in I^2 R losses in their network - which is power they have to pay for but can't sell.
Then long along comes a load of embedded generation (particularly solar PV in residential areas) which change the current distributions - including potentially reversing voltage drops on sunny days - and they find themselves with complaints of inverters shutting down due to over-voltage.

Will be interesting what other non-linear loads are disappearing from the grid , like incandescent lamps that over ran and shortened their lives (esp when rated 240 not 250V).
Lower wattage hoover motors ! ?

If you are dealing with a fixed shaft power output then yes, the V x I should be more or less constant. But depending on what a motor is driving and the specific characteristics of the motor the current may well go up with voltage.

Typically AC motors are more-or-less constant speed, so they should broadly show the "V x I = constant" sort of characteristic, but universal motors (and DC motors driven from a simple rectifier arrangement) typically vary speed proportionally to voltage so if you have a load like a fan where the torque/power goes up a lot with speed then they would draw more current as the voltage goes up (due to delivering more shaft power).

Motor current can increase as the voltage falls, but that is not because it is inductive.

A pure inductance is linear and passes a current proportional to voltage. I=V/xC = V/2.f.L The nearest approximation to a pure inductance found in normal electrical work would the be the primary of an unloaded transformer. Within reason, the magnetising current is proportional to the voltage, but as it lags the voltage by 90° it absorbs no power. In any real transformer there are losses even when unloaded so there is also a power component of the current which, although it increases with increasing voltage, is non-linear as it depends on the magnetic characteristics of the iron core which are also non-linear.

Motors much more complicated; different kinds of motors with different kinds of loads behave in very different ways. To expand on PC1966's comment about motors in which the speed is determined mainly by the frequency (synchronous, induction) and those where the torque is nearly constant, consider a hoist, where the torque depends on the weight being lifted, not the speed it is moving. The power component of the current is approximately proportional to torque, but inversely proportional to flux density (from the general motor characteristics.) Since flux density increases with voltage (although not linearly) the effect of voltage drop in the cable would be to increase the power component of current but decrease the wattless magnetising component. The power factor would improve and there would be some arbitrary variation in current magnitude.

As he points out, a motor of non-constant speed, especially when coupled to a load with a steep torque/speed curve, is likely to show decreased power and current due to voltage drop. Therefore it is impossible to make the generalisation: 'With increasing voltage, motor current varies as...." For a more thorough explanation we would have to delve into some AC circuit maths and get into motor characteristic equations etc.

Motor current can increase as the voltage falls, but that is not because it is inductive.

A pure inductance is linear and passes a current proportional to voltage. I=V/xC = V/2.f.L The nearest approximation to a pure inductance found in normal electrical work would the be the primary of an unloaded transformer. Within reason, the magnetising current is proportional to the voltage, but as it lags the voltage by 90° it absorbs no power. In any real transformer there are losses even when unloaded so there is also a power component of the current which, although it increases with increasing voltage, is non-linear as it depends on the magnetic characteristics of the iron core which are also non-linear.

Motors much more complicated; different kinds of motors with different kinds of loads behave in very different ways. To expand on PC1966's comment about motors in which the speed is determined mainly by the frequency (synchronous, induction) and those where the torque is nearly constant, consider a hoist, where the torque depends on the weight being lifted, not the speed it is moving. The power component of the current is approximately proportional to torque, but inversely proportional to flux density (from the general motor characteristics.) Since flux density increases with voltage (although not linearly) the effect of voltage drop in the cable would be to increase the power component of current but decrease the wattless magnetising component. The power factor would improve and there would be some arbitrary variation in current magnitude.

As he points out, a motor of non-constant speed, especially when coupled to a load with a steep torque/speed curve, is likely to show decreased power and current due to voltage drop. Therefore it is impossible to make the generalisation: 'With increasing voltage, motor current varies as...." For a more thorough explanation we would have to delve into some AC circuit maths and get into motor characteristic equations etc.

It would need a "constant power" load - but most loads are speed dependent, reducing power when speed reduces.

Most pumps and fans reduce torque considerably with reducing speed. Something like a hoist would take a fixed torque (almost) independent of speed.

The only thing I can think of with a negative speed-torque curve would be a variable output pump or compressor - configured in constant pressure mode. I've an idea some hydrovane compressors are like that, and a variable output hydraulic pump certainly could be. I've a feeling that a refrigeration compressor would also fit the need (assuming constant cooling load) - but that's a complex system to get your head around and I've had a long day
So you've a hydraulic pump or air compressor configured to maintain system pressure by varying pump output (e.g. swashplate hydraulic pump). If the speed drops, then for a given load, the pump will need to pump more fluid per revolution. Shaft power would remain roughly constant, so for a reduced input voltage, the current would increase - slightly more than needed to maintain constant power as there would be more heating in the motor.

An excellent example there (the variable-displacement pump) of a load with a backward slope, much steeper than that of the motor itself. I don't have much time but I would like to put a selection of squirrel cage motors in real applications on a 3-phase variac and plot a few points. I'm sure I've done it in the lab and seen the current rise with the rotor magnetic load angle on a synchronous motor as the stator volts drop under constant excitation, but not sure if I've done the same with inductions.

What is worth remembering, for the OP, is that in real-life situations the voltage drop on a practical circuit is a small fraction of the supply voltage. Regardless of the exact value of the effective dynamic resistance or load curve, the current variation will also be modest.

Ooh, electrical machines labs - that takes me back ... a loooong time ! I can just about recall doing experiments with a (small) synchronous motor - loading it and watching the pen move on the plotter, until the load goes too high and the pen does abstract artwork all over the sheet as the motor loses sync For some reason, I can't possibly think why, we were never allowed to play on the bigger machines, try syncing a generator with the 3 lamp technique, and stuff like that - done on real LIVE equipment connected to the mains I bet H&S have watered things down a bit since.
But at least we were mostly spared the need to understand the amplidyne and metadyne that were also in the lab. These modern VSDs take so much of the "fun" out of things.
It was also during my time at uni that the lab got rid of the old battery system for DC stuff. Room full of large cells, with a 'kin big switch and plug board that allowed us to pick off anything from (I think) 2V up to 240V from the battery and route it to the terminals on the bench. With directions to check what voltage is there before connecting anything - someone else might have been doing a different experiment. Of ocurse there were temptations to try "unauthorised experiments", but I think our year wasn't all that adventurous compared to some of our predecessors They got rid of that as the cells were failing and it was going to cost a fortune to replace them - so they just got a large transformer/rectifier to run the larger DC machines and bench PSUs for the smaller stuff.

Now I'm getting all nostalgic when I think how much more I'd learn if I could go back and have another go - with the benefit of a bit more knowledge to start with.

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