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Over time, a larger proportion of electronics are constant-power loads, while they used to be constant-resistance. Will this lead to an unstable grid? If not, why not?

Constant power loads at home are battery chargers, TVs, many appliances.

In a constant-resistance environment, the grid is self-regulating to a certain extent, as if power generation cannot match power consumption, then a slight drop in voltage would reduce the consumption.

In a constant power environment, this self-regulation would not exist, as a slight drop in generator voltage would just lead to an increased current draw and a constant power consumption. This sounds like it would lead to a run-away process if left unchecked.

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  • \$\begingroup\$ Related: Producers consumers balance in the grid \$\endgroup\$ – Tyler Oct 27 '19 at 22:24
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    \$\begingroup\$ Not an answer to the core question about the effect of constant-power loads, but all these loads are not that big amount compared to other loads. See epa.gov/energy/electricity-customers - appliances and electronics are not the dominant consumer, and even for that washing machines, dryers, refrigerators and ovens totally outclass TVs and battery chargers. Boiling water for a single cup of coffee or tea takes more energy than fully charging your phone. \$\endgroup\$ – Peteris Oct 27 '19 at 23:04
  • \$\begingroup\$ @Peteris Datapoint: 1 kWh = 850 litre-degrees-C for water. So rising say 1 litre (4 standard cups) of water from say 15C to 100C = 85c x 1 litre = 85 litre.degrees = 0.1 kWh = 100 Wh. Charging a phone battery of say 3.3 Ah x 5V from supply at say 80% overall efficient (typically TWO smps in series) = about 20 Wh. So yes, even an iPad 10 Ah battery takes less than a kettle boil. \$\endgroup\$ – Russell McMahon Oct 28 '19 at 0:59
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Yes, potentially a lot of constant-power loads will destabilize a feedback loop that keeps the grid voltage steady. This is because constant power is symbolized by $$power = V \times I = constant$$ while resistive (like a heater) loads are symbolized by $$power = V \times I = V^2 \times constant$$

The grid must deliver extra power whenever it sees a voltage dip. The heater load makes a V versus I function that has a strong power decrease when V goes low, i.e. it helps the grid feedback control (lower power load is like higher power supply). The constant-power load, however, has a proportional current-load INCREASE when V goes low, which means it requires a larger current increase than the voltage deficiency alone would indicate. It is thus fighting the power-grid voltage regulation mechanism.

Fighting is normal, but you don't want the power-grid to lose that fight.

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  • \$\begingroup\$ Info only: Different scenario - same result. A DC feed to Sardinia ended up in instability with the tail wagging the dog due to load-supply dynamics. Long ago. \$\endgroup\$ – Russell McMahon Oct 28 '19 at 1:01

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