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Can anybody help explain in simple terms how adding capacitors to a 120/240/higher AC power circuit helps improve power factor without blowing up the capacitor? I'm familiar with capacitors in DC circuits as tanks for storing energy but it seems like that concept collides with the idea of putting a capacitor in parallel with AC voltage in a circuit - wouldn't the constant reversal of polarity destroy the capacitor? And wouldn't it be a huge waste of energy (every time the capacitor charges in one polarity it discharges and works against the AC line voltage as it swings to the next opposite polarity...)?

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    \$\begingroup\$ What exactly do you think should blow up the cap? And what mechanism should waste energy? Both only happen when resistance converts power to heat, which can easily minimised by using caps with low enough ESR \$\endgroup\$ – PlasmaHH Jun 13 '17 at 20:11
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    \$\begingroup\$ @PlasmaHH I imagine taking a DC power source and charging a large electrolytic capacitor. Then switch the polarity on the capacitor and marvel at the giant spark, then do that again, and again, 120 times per second. That was what I imagined happening when AC is applied to a capacitor. However I now realize that sizing the capacitor appropriately plays a significant role in power factor correction. Much smaller capacitance values are used for this than would be used for DC filtering (which might be say 22,000uf). \$\endgroup\$ – Brad Hein Jun 30 '17 at 20:01
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    \$\begingroup\$ the same applies here: when the ESR is low enough, minimal power is dissipated in the cap \$\endgroup\$ – PlasmaHH Jun 30 '17 at 20:52
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I'm familiar with capacitors in DC circuits as tanks for storing energy

Don't think about storing energy when talking about capacitors in AC circuits!

wouldn't the constant reversal of polarity destroy the capacitor?

As far as you don't have a polarized capacitor: Normally no.

It depends on the waveform (e.g. square or sine) you apply to the capacitor. We are talking about sine voltages here.

Unlike batteries capacitors can be charged and discharged without wearing.

And wouldn't it be a huge waste of energy

Indeed you have some energy loss because of current flow in the capacitor. However due to the better power factor you have less current flow in the power supply system.

This means you also have less energy loss in the power supply system.

You pay a minimum of more energy loss (in the capacitor) and get rid of a much higher energy loss (in the supply system).

how adding capacitors to ... AC power circuit helps improve power factor

Many AC engines behave like they have a coil in parallel.

Just like capacitors coils store energy and emit that energy later. When operating at 50Hz for example (sine wave voltages) both capacitors and coils store energy 100 times per second and in between they emit the energy stored before 100 times per second.

For the power supply system this is not good:

The energy stored in the coils must be transported from the power station to the engine over transmission lines and when the energy is emitted it must be transported back. Transmission lines have a resistance (of course) so whenever energy is transported over the transmission line there is also a loss of energy.

Therefore it is not wanted that electrical devices behave like they have a coil in parallel but unfortunately for many devices this effect cannot be avoided.

The more detailed explaination how the capacitor works would be looking at the differential equations of the capacitor and the coil.

The more simple explaination is looking on the way the two parts are storing the energy:

The higher the absolute voltage is the more energy is stored in a capacitor. When the voltage is zero the capacitor has no energy stored. This is true for DC, AC and any other voltages.

When a coil is connected to a sine voltage (note: this is only true for sine voltages) it is exactly the opposite: When the voltage is in the maximum the coil does not have any energy stored; when the voltage is zero the coil has the maximum of energy stored.

This means: Whenever the coil emits energy the capacitor will store energy and whenever the coil stores energy the capacitor will emit energy.

The trick is to choose a capacitor whose capacitance has a value so that the energy stored by the capacitor and the energy stored by the coil is the same.

In this case no more (additional) energy must be transported over the transmission lines because the energy is exchanged between the coil and the capacitor.

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  • \$\begingroup\$ So do you need to use special AC capacitors then? I've only really used electrolytics and they explode when connected backwards so I don't understand how they'd work with AC, even if it is a sine wave! (Sorry I'm a bit confused about what you meant by "as far as you don't have a polarized capacitor" - not sure whether this means the polarization is unimportant or that you don't consider using polarized ones.) \$\endgroup\$ – Malvineous Jul 18 '18 at 2:16
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    \$\begingroup\$ @Malvineous There are capacitor types that have no polarity and types that have a polarity (such as normal electrolyte capacitors). You have to use some type that has no polarity. So normal electrolyte capacitors indeed cannot be used. \$\endgroup\$ – Martin Rosenau Jul 18 '18 at 4:30
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Can anybody help explain in simple terms how adding capacitors to a 120/240/higher AC power circuit helps improve power factor without blowing up the capacitor?

Sure.

  • wouldn't the constant reversal of polarity destroy the capacitor?

No. In a capacitor the voltage and current are 90° out of phase so the power dissipated in them is zero (in the ideal case). In real capacitors there is a little internal resistance and an internal heating caused by movement of any mobile charges in the dielectric. The power-factor correction capacitor design will take this into account, be rated for the mains frequency and should remain inside their safe working temperature in continuous use.

If there are excessive harmonics in the current the situation can change. e.g., If there was a high 13th harmonic at 650 Hz at, say, 10% of the 50 Hz current value you might find that the heating effect of the harmonics was more severe than the 50 Hz and the capacitors would fail early.

And wouldn't it be a huge waste of energy (every time the capacitor charges in one polarity it discharges and works against the AC line voltage as it swings to the next opposite polarity...)?

Not when set up properly. Remember that the reason for installing them is to counteract the inductive current in the load. The result is an overall reduction of reactive current towards zero.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. An inductive load (L) without capacitive correction would end up with a kVA magnitude greater than the kW magnitude. Adding the equivalent capacitance will cancel it out.

In effect we set up a tuned circuit between the L and C so that the reactive current component sloshes around between them without drawing additional current from the supply. Without them the out-of-phase current causes a higher current to be drawn from the grid than that which is required. This causes increased voltage drop along the supply lines and this is where the majority of power loss occurs.

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  • \$\begingroup\$ If I could accept two answers I would also accept this excellent description. Combined with the previous answer I realized I had some misconceptions. The part about the current sloshing between load and L-C tuned circuit instead of supply line was quite helpful as well. \$\endgroup\$ – Brad Hein Jun 15 '17 at 3:29

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