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If two points in circuit are not connected there is supposed to unlimited resistance between them. So when the compressor is added to circuit it is either very high resistance or very low. In either case the voltage is not supposed to fluctuate but adjust according to distribution of voltage across resistance according to ohm's law. Does it mean that it sinks current flow? or what is it that happens?

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    \$\begingroup\$ The title makes sense, but the rest is gibberish. \$\endgroup\$ – Olin Lathrop Jan 2 '17 at 16:35
  • \$\begingroup\$ All the outlets are connected at the fusebox or daisy chained around the room or both, so (unless you have 3ph power) every outlet is in parallel and shares the same supply which is why the lights dim when the compressor starts up, they're sharing the same supply and so any disturbance at one outlet will affect everything else in the house (although most appliances won't mind too much) \$\endgroup\$ – Sam Jan 2 '17 at 21:05
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I can't understand the text of your question, so I'll just answer the question in the title.

While it's true that the power supplied by the utility can generally be considered a voltage source, in practice, the wiring always has a small but nonzero resistance. This, combined with the very high current drawn by a large motor at startup, creates a voltage drop that is large enough to notice on parallel loads, including light bulbs.

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An AC motor has a very low resistance when it is initially connected to power. As the speed increases, the resistance increases to a higher value. That means that the current is initially very high then decreases to a lower value. The high current causes a voltage drop in the wiring between the source and the motor. The lights or anything else that shares the wiring that carry the high current experience a temporary voltage drop. The starting current of a motor can exceed the current that the circuit can carry continuously. If the load would jam, the bearings fail, or anything else prevent the motor from starting quickly, a circuit breaker or fuse would open. All of that is generally normal operation, but the dimming lights could indicate that the wire size is smaller than it should be for the distance between the source and the motor.

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  • \$\begingroup\$ Actually, the resistance doesn't change at all. As the motor speeds up, the "back EMF" that it produces goes up. This reduces the net voltage across the motor's resistance, and this is why the current drops at speed. \$\endgroup\$ – Dave Tweed Jan 2 '17 at 16:50
  • \$\begingroup\$ I am assuming this is an induction motor. As such, the Steinmetz circuit is the usual model. \$\endgroup\$ – Charles Cowie Jan 2 '17 at 17:57

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