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flatulent1 | 14 years ago
Uh, no. Most of the voltage drop in the source occurs in the resistance of the wiring, and you didn't even figure resistance into the equation. Intuitively, the surge current more closely matches that of a capacitor charging through resistances. It isn't even a matter of the motor inductance or the frequency of the power. In fact the early surge would be seen even if we used D.C. power and a D.C. motor. The time constant matching the light dimming involves the mass of the rotor being brought up to speed. The load is actually more resistive at startup than when an A.C. motor is up to speed and coasting.
A pure inductor starts at zero current the instant voltage is applied and it ramps up at a rate determined by the inductance, at D.C. or very low frequencies rising to where it becomes limited by the series resistance. It's pure capacitors that draw a maximum current at the instant voltage is applied to the circuit.
I believe that many people get lost in the math and lose a feel for what is happening. What you said is completely backwards from reality. Mechanical analogies are far easier for people to get a feel for than Laplace Transforms. Compare the behavior of an RLC circuit to a machanical system with a shock absorber, spring, and a mass.
People may get confused over a critically damped circuit, but the idea of a car oscillating up and down after a bump with no shock absorber is easy to grasp.
mturmon|14 years ago
Here's a motor's equivalent circuit from Wiki:
http://en.wikipedia.org/wiki/Induction_motor#Equivalent_circ...
This does have lumped resistance elements, but by the time you put the two inductors and two resistors in to the differential equation with forcing
you're not going to be in a good place as far as intuition goes.In fact, overall, I'm not seeing much intuitive sense in your summary. Once you really delve in to it, the subject of starting electric motors (AC and DC) is very complex.