I have vague memories of the overlap between physic and electrical engineering, and saying a voltage sag increases the current seems a bit odd, at least at first.
Then I remembered we're talking AC, and things like refrigeration and air-conditioning involve electric motors, and there are phase shifts, and my poor brain melts into a puddle on the floor.
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timeasmymeasure
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Date: 2016-06-12 12:15 pm (UTC)Then I remembered we're talking AC, and things like refrigeration and air-conditioning involve electric motors, and there are phase shifts, and my poor brain melts into a puddle on the floor.
But electric motors draw a lot more current when they start, which means that the balance between the Thévenin resistance/reactance of the utility and the resistance/reactance of the house shifts. At this point Google produces a very startled rabbit to tell me that the RTH is about 0.25Ω which means the house has to be around 1Ω when that 100A breaker tripped. If the open circuit voltage is 120V, the voltage across the terminals will have dropped to around 100V.
It looks to me that the rise in current is causing the voltage sag, not the other way around.
Of course, the utility isn't a simple, fixed, Thévenin equivalent, but I think the model will hold over the few seconds surrounding a motor-start. And if the VTH has dropped, a starting motor might draw more current for longer, a bit less power to accelerate, and circuit breakers react to current and time. I remember seeing some figures, back when I was working on a 3-phase motor, and the star-delta starter switch exploited the short-period overload limits.
Yes, I saw it dim the streetlighting.