# Will a large amount of constant-power loads make an electrical grid unstable?

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.

• – Tyler Oct 27 '19 at 22:24
• 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. – Peteris Oct 27 '19 at 23:04
• @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. – Russell McMahon Oct 28 '19 at 0:59

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 $$power = V \times I = V^2 \times constant$$