Voltage drop.

[Jako13!]

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

 

[Wilko]

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.

 

[telectrix]

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. ).

 

[plugsandsparks]

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

 

[Lucien Nunes]

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.

 

[Jako13!]

Thanks guys.

So is my understanding right.

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?

 

[telectrix]

Thanks guys.

So is my understanding right.

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.

 

[Lucien Nunes]

so will the current causing the overall power to also be reduced?

Revisit my post #5 above. Sometimes it decreases, sometimes stays nearly constant, sometimes increases, depending on the nature of the load.

 

[Jako13!]

Revisit my post #5 above. Sometimes it decreases, sometimes stays nearly constant, sometimes increases, depending on the nature of the load.

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.

Am I right. Thank you for your patience guys.

 

[pc1966]

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).