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Let's say I'm using a 100% efficient battery (AC battery, so there's an inverter in front of it) to power a motor with a power factor of 0.5. Let's also assume the inverter cannot correct the power factor for some reason (normally they can, right?) and is 100% efficient. Let's also assume 0 resistance in the power lines.

If I hook up the battery to a 1 HP motor with 100% efficiency, will the battery start losing more than 1 HP? I would assume not since the inverter's job is just to convert power from one form (DC) to another (AC) and it doesn't take power to have current with 0 voltage, correct (this would happen every cycle because the voltage and current wave form are out of sync due to the bad power factor)?

The only concern I have is that batteries have a limited amount of current, so if current was always being extracted from the battery (even at 0 voltage due to a bad power factor), the battery would still be losing potential power.

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  • \$\begingroup\$ The power factor will mean that the inverter would have to be rated for at least 1.5 kVA (i.e. the apparent power of the load: 1 HP / 0.5) \$\endgroup\$
    – τεκ
    Oct 16, 2017 at 22:25
  • \$\begingroup\$ Yeah, let's say I'm dealing with an inverter that isn't able to handle power factors other than 1 for some reason, so it's going to be affected by the motor's bad power factor (motor will need more current than it should with a good power factor). If that makes sense I mean. I know it's a weird question because normally the inverter would fix the power factor somehow (how does it do this? By shifting the current wave form or the voltage wave form)? \$\endgroup\$
    – MechE
    Oct 16, 2017 at 22:37

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The diagram below shows a basic inverter power circuit with a battery source and induction motor load. An induction motor is an inductive load because it has a magnetic field that must be maintained for it to operate. Since the magnetic field changes polarity every time the current reverses, it must constant exchange stored energy with the source. That circulating energy result in a lagging power factor. It is not a "bad" power factor, it is the power factor that the motor requires for its operation. The only energy that is used as a result of the lagging power factor is the losses in the conductors carrying the current.

If the circulating energy would be circulated between the battery and the motor, that would involve the motor adding to the battery charge during part of each cycle of the AC waveform. That might be possible, but it would add to heating and the wear-out mechanism of the battery. To avoid that, the inverter should have a diode to prevent current from flowing back to the battery. A capacitor is added to exchange stored energy with the motor.

Energy is returned from the motor to the capacitor through the diodes connected inversely parallel to the transistors. The result is a large "AC" ripple current added to the DC current. The current to the inverter has a DC component supplied by the battery and an AC component that flows in the capacitor.

If the capacitor and input diode are omitted, the battery is likely to fail or have a shorter than normal life. If the capacitor is not large enough it may fail or have a shorter than normal life. In either case, the motor may have difficulty developing rated torque, particularly starting torque.

enter image description here

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  • \$\begingroup\$ Okay, so if a battery was transferring power to a motor in AC form and the inverter wasn't needed somehow, the battery wouldn't be losing amp hours at the end of the day because although current would sometimes leave the battery at 0 volts, current would also enter the battery at 0 volts. This might wear out the battery's mechanisms though. So moving to a more "real" case, since an inverter would need to be in front of the battery to actually output AC power, would the battery lose additional current trying to get the inverter to output the extra current needed to run the 0.5 PF motor? \$\endgroup\$
    – MechE
    Oct 17, 2017 at 18:52
  • \$\begingroup\$ The inverter would have losses. The losses might be as small as 5% of the motor power for a 1 Hp motor. Losses would be higher for an inverter that boosts the voltage from a low battery voltage, perhaps 10% or 15%. The losses associated with the power factor would be more like 0.5%. \$\endgroup\$
    – user80875
    Oct 18, 2017 at 20:14

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