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Why doesn't a motor slow down when a capacitor is connected to it in parallel? Here is my understanding. During the first half cycle, the capacitor charges up while the load is getting power. During the zero point, no power from the source goes to the load and the capacitor is holding a charge, but not charging. During the second half of the cycle, the load gets power in reverse and the capacitor starts to discharge in the opposite direction. So there should be two overlapping waves one wgoing go the motor and the other leaving it.

Essentially, the motor would slow down or stall because it is getting negative voltage, but it doesn't. Very confused and please no MATH in your answers. It will only frustrate me further.

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  • \$\begingroup\$ A motor is low impedance at start from winding resistance. For a capacitor to conduct the 8x surge currents that a motor uses for a short circuit stop ( lawn e-mower) it would use the same current as it starts up it might have to be as big as the motor and keep increasing in size. However to improve the power factor on induction motors some do use starter caps to help reduce the inductive shifted currents of the starter winding. \$\endgroup\$ Oct 5 '16 at 1:40
  • \$\begingroup\$ @Tony, Thanks, but is my understanding correct? I also forgot that there is a third wave form, back emf. \$\endgroup\$
    – user148298
    Oct 5 '16 at 1:46
  • \$\begingroup\$ Torque is only due to current and the cap current is low due to high impedance for the small values available for high AC line voltage \$\endgroup\$ Oct 5 '16 at 1:55
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    \$\begingroup\$ Your voltage source is stiff so it does not matter what you connect in paralell to it. Apply KVL. \$\endgroup\$
    – winny
    Oct 5 '16 at 7:31
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During the zero point, no power from the source goes to the load and the capacitor is holding a charge, but not charging.

This is your misunderstanding.

Since the capacitor is directly across the mains with very low resistance anywhere in the circuit the capacitor voltage will follow the mains exactly in sync. There will be no lag (as there is in an RC circuit, for example) because there is no resistance.

The capacitor has no affect on the operation of the motor.

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You need to think of the motor having two parts. One part is an inductor or electromagnet that provides a magnetic field but doesn't burn up any power. The other part takes power from the source and transmits it as mechanical power to whatever the motor is driving. The electromagnetic accumulates energy from the capacitor during one half cycle and returns it during the next half cycle. The electromagnet and the capacitor can be thought of as being completely separate from the source and the operation of the motor. You can think of power going directly from the source to the rotor of the motor where it is converted to mechanical power and delivered to the load. There are losses in the electromagnet, the rotor and in the capacitor. You can think of power coming from the source and being converted to heat in those components.

Induction motor operation is usually analyzed without considering back emf. A very simplified equivalent circuit only includes the electromagnet and a variable resistor. The variable resistor represents power converted to mechanical power and some of the losses.

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  • \$\begingroup\$ What about back emf? Where does it go? Dirty power? back to the capacitor? \$\endgroup\$
    – user148298
    Oct 5 '16 at 1:51
  • \$\begingroup\$ Yikes. I was taking abut induction motors. EE folks sometimes use AC and induction interchangeably adding to more com fusion. \$\endgroup\$
    – user148298
    Oct 5 '16 at 1:55
  • \$\begingroup\$ Yes. An induction motor is what I am talking about. See revised answer. \$\endgroup\$ Oct 5 '16 at 2:10
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To a first approximation, the motor is irrelevant to the capacitor, and the capacitor is irrelevant to the motor.

The reason this is true is that the AC line is intended to be an AC voltage source (and in fact what you have drawn is literally one) - up until the point where you break something, you can to a first order approximation consider that the voltage on the line terminals is the same no matter what you connect.

Of course while this is absolutely true for what you have drawn, there is a degree to which this is not strictly true for actual machinery connected to an actual AC line of non-zero source impedance.

For one, the capacitance will be an admittance at AC, which is to say that it will load the line during part of the cycle and so (assuming finite wiring resistance) both reduce the voltage and shift the phase a little. But the motor will only see a different phase - which it doesn't care about at all and a slightly lower voltage, which if an induction motor not critically loaded it doesn't particularly care about either, since the speed absent substantially loaded slip is determined by the line frequency.

The practical reason why someone would actually build the circuit drawn has nothing to do with speed control, but rather with a desire to balance the inductive reactance of the motor with local capacitance, so that the load to the AC grid is more nearly resistive. Although neither inductive or capacitive reactance consume real power, the cyclic AC power in and out of them has to flow through the real resistance of the distribution wiring where it does dissipate real power as heat, so industrial customers are billed a surchage for reactive loads, and may find it desirable to locally compensate out the reactance.

So if you want a model where something actually happens, you need to insert a resistor between the motor / capacitor assembly and the AC source, to model line losses.

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