What determines the current flow in a load?

Question

If we apply a load of 100KVA to a 100 volts source, we will get 1000 amperes of current through it.

If we apply the same load of 100KVA to a 1000 volts source we will get 100 amperes of current in the load.

If voltage is high load will draw less current and if the voltage is low load will draw more current.

How does the load do that? What is in the load that draws the current according to the voltage? The load itself is not a living being so how does the load have that sense to draw current according to voltage?

Answer 1

You can buy a programmable load. Most are intelligently controlled, and work by measuring the applied voltage, then switching load elements in or out to draw an appropriate current. I used one that had 'constant resistance', 'constant current' and 'constant power' settings. It was even fast enough to work with a mains AC waveform, and program the current to mimic a limited range of non-unity power factors. Such a load would be able to draw constant power, if it had been designed to work over the required range of voltage and current.

Most loads that you meet are not like this however. Constant resistance loads will tend to draw power as the square of the applied voltage. Loads with a positive temperature coefficient, metal filament lamps for instance, will tend to draw a more-nearly constant current, as their resistance goes up with the temperature. Some loads like electric arcs actually have a negative voltage coefficient, so must be stablised by an external impedance or controller.

Answer 2

How does the load do that? What is in the load that draws the current according to the voltage? The load itself is not a living being so how does the load have that sense to draw current according to voltage?

The load you are speaking of is an ideal constant-power load. It’s a theoretical construct used in circuit theory. The theory doesn’t care how such a load is implemented, because the entire specification for the load is its behavior – for the behavior is the only thing you observe, and such an ideal load has no internal state that you could inspect. It’s a “black box” with terminals.

In practice, only approximations of such loads exist, and they typically require some feedback control.

For example, the load could be a heater with a thermostat. When looked at through a sliding time average, the power consumed by that load will be equal to the heat loss rate, and not depend on the supply voltage, as long as the heater has low enough resistance.

This is actually observed in electrical power distribution and such loads have negative resistance. There are so many electricity consumers that their aggregate behavior is self-averaging. The discrete changes of state of each load device (eg. a switch flipped on/off) are not individually observable, but rather the load follows the number of devices, the demand for heat, etc., as if they were continuous variables.

Consider a residential area where everyone has electric heaters with thermostats controlling the temperature. The higher the voltage, the fewer thermostats (on average) are turned on at any given time, since the heaters will generate same amount of heat in a shorter time. Thus the higher the voltage, the lower the current, when the power needed to do the heating job is “fixed” by weather conditions and social traditions (eg. holiday cooking or taking morning showers).

In other words, taking a positive resistance and adding an active control mechanism can transform it into a negative resistance, even if the control mechanism is as crude as a mechanical thermostat – as long as there’s enough of such crude controllers to average the error out.

Answer 3

You are specifying (pre-determining) the power (VA) to be consumed by the load. So if you do the math the VA value can be satisfied by either 100V x 1000A = 100KVA, or by 1000V x 100A = 100KVA, (same result). So essentially you determined the current flow (at a given voltage) when you specified the VA value.

Answer 4

One thing that approximates a constant power load is a switch mode power supply with a fixed load on the output. The power supply converts variable voltage input power to fixed voltage output power and supplies that to a load.

For example, if you had a buck converter that puts out 100 V, and you connect a 0.11 Ohm resistor to it, it will put out 900 Amps. So it is delivering about 90 kVA to the resistor. If it also dissipates 10 kVA in lost power, that means it will consume 100 kVA from the input supply.

So our theoretical buck converter with a resistor as a load on the output is similar to a constant power load.

If I supply the buck converter with 1000 V on the input, the output will still be 90 kVA. The input power will still be about 100 kVA.

In reality, though, buck converters that can accept a 10:1 input voltage range are unusual. And even if they can accept such a wide range, the efficiency will usually not be consistent across the full input voltage range. Generally efficiency tends to go up as Vin is closer to Vout. So the buck converter approximates a constant power load, but has some variation of power with voltage.

Also, 100 kVA is a high power level for a buck converter. Certainly I have zero experience designing something like that. I suspect it is possible, but it is outside my experience.

There is also something called an electronic load or programmable load. The ones I have seen are usually lab bench units. They have different modes of operation, but if you set it to act as a constant power load, it will monitor input voltage and current, and adjust the input current to maintain the target power level. Again, given the 10:1 voltage and current ratio on the question, you would likely need a massively over-sized electronic load (rated at much more than 100 kW).

I am sure there are applications for 100 kW programmable loads, but they would be somewhat specialized. You probably can't find a price list with model numbers and just order it online. You would need to contact the supplier and discuss your needs to work out the details.

In both of these cases, the way the load would work is by using current control in conjunction with sensing of voltage and/or current at the input and/or output of the module. It would not be just a "dumb" load.

There are probably other ways to approximate a constant power load, but they are all going to involve some type of feedback so that the input current naturally goes down as the voltage goes up.

Answer 5

I think you may have seen your third paragraph "If voltage is high load will draw less current and if the voltage is low load will draw more current." in a discussion of AC power distribution systems.

In those systems, transformers are used to convert between voltages, and a transformer passes power (voltage times current). The primary power into a transformer is equal to the secondary power out of the transformer (neglecting losses).

For your example of a 100 KVA load requiring 100 volts, the load will draw 1000 amp. If the electric company supplies power to you through a 10:1 tranformer, so they can deliver 1000 volts, their side of the transformer will only draw 100 amp.