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Let me explain you why I am asking this question.

The other day I was studying about power factor correction of a (step up or any) transformer. It said that on the output side of transformer's secondary, when the voltage is higher, there's no current flowing because of highest impedance, and when voltage starts to reverse, the impedance starts going down and when voltage is zero, the current becomes highest. So voltage and current are out of phase, when voltage exists, current doesn't exist, and when current exists, voltage doesn't exist.

So what my understanding of PF correction with caps is that, when load is connected to the output of a transformer's secondary along with cap in parallel, the capacitor gets charged when the voltage is higher and current is lower or zero, and when the voltage in secondary drops, the current from secondary goes directly to load because cap in parallel is already fully or 63% charged, at that time when voltage in secondary is zero, the charged capacitor delivers the voltage to the load (and why not back to secondary?) and this way the load receives the voltage and current at the same time and it consumes most of the available power.

So by my study, what I understand is for PF correction, the capacitor gets charged only wiht voltage and no current. However what I also studied is that there has to be current flowing in order to charge a capacitor. That's why I am confused and hereby asking whether capacitor gets charged without current and only with voltage, if they do, then how does actually capacitor corrects power factor?

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    $\begingroup$ "voltage is higher and current is lower" is talking about the current going to the load (e.g. motor). By charging when the current is too low it raises the current and improves the powre factor $\endgroup$
    – user253751
    Mar 27 at 10:55

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So voltage and current are out of phase,

Yes, but ...

when voltage exists, current doesn't exist, and when current exists, voltage doesn't exist.

... no.

The definition of capacitance is that the electrical potential is proportional to the charge,

$$ Q=CV $$

which you can relate to the current by differentiating both sides:

$$ I= C\frac{\mathrm dV}{\mathrm dt} $$

This isn't a statement that one of these things causes the other. They're just the same. The only way to change the energy per charge (i.e. the voltage) across a capacitor is to change the charge stored in it. The flowing charge (i.e. the current) is proportional to the rate of change of the voltage, because the charge and the voltage are proportional to each other.

It's not correct that current and voltage are mutually exclusive in a capacitor. In an alternating-current circuit, there are only two brief instants where the current is zero: while it is changing sign. At all other times, there is non-zero current and voltage across the capacitor.

Power-factor corrections are about synchronizing the voltage and the current waveforms, the way that those waveforms are synchronized in a purely resistive circuit. If you force an alternating-current waveform through a capacitor and measure both the current and the voltage, you'll find that the maximum positive current happens shortly before the maximum positive voltage. We say that the current "leads" the voltage. That's different from an inductor, whose equation

$$ V = - L\frac{\mathrm dI}{\mathrm dt} $$

means that the current "lags" the voltage. By cleverly combining capacitive and inductive elements in our transformer, we can get the "lead" and the "lag" to cancel out.

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  • $\begingroup$ So if current and voltage out of phase doesn't mean that when voltage exists current doesn't exist. So what does the scope shot of a transformer mean which shows that when voltage has highest value, the current wave is at zero and vice versa? $\endgroup$
    – Yogie
    Mar 26 at 17:01
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    $\begingroup$ Those are the "brief instants." Any finite time interval has nonzero $V^2$ and nonzero $I^2$. $\endgroup$
    – rob ♦
    Mar 26 at 17:12
  • $\begingroup$ To add one more besides the AC circuit example, I remember in a lab exercise where we measured the resistance in a voltmeter. We hooked it up to a charged capacitor (with known / assumed capacitance), and watched as the voltage dropped, slowly but surely over minutes, as the capacitor discharged through the voltmeter. Clearly there is both current and voltage at the same time. $\endgroup$
    – Arthur
    Mar 27 at 8:51
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The constitutive relationship of a capacitor is, at any instant of time:

$$V = \dfrac{Q}{C}$$

Where $C$ is a property of the structure and does not change. This relationship tells you that there is no way to change the $V$ across a capacitor, if not by altering the charge $Q$ on one of its plates.

Now, we normally assume that at the beginning of our time, the plates of the cap are neutral. They have no charge imbalance (so the $Q$ in previous formula is actually zero at the beginning of our observation).

Therefore, any charge imbalance accumulating on the plates of the cap must correspond to a charge travelling, in the circuit around the cap, into the 'positive' plate and out of the 'negative' plate. This is, of course, a current.

Therefore, there is no way you can change the voltage across the plates of a cap, without pouring charge on it and therefore having a current across its volume. In lumped elements theory, since charge cannot accumulate in any volume (and the cap is considered to have zero volume), the charge entering on one end must exit the other end, you have your classic current definition abiding to Kirchoff's law and all the rest.

This is the first part of the question.

On the second part, about the power factor correction. I would tend to simplify the scenario to a generator trying to deliver power to a (resistive) load and this load has an inductor in parallel. While the active, or average power is tied to the only portion of energy that's leaving the system, that is the one going into the resistor and therefore converted to some other form, there still is current flowing into the inductor and voltage developed across it which while do not 'leave' the system (because they're stored in the generated magnetic field), can cause stress to the circuit. Therefore, in sinusoidal regime the concept of apparent power is used to take this kind of "stress" into account, although as said the portion of real, or average power flow out of the circuit is a fraction of it.

PF measures this ratio: the higher, the better because the kind of power we're stressing the system with (the apparent power), is also the power we're delivering. So we're not wasting anything.

This is achieved by resonating the inductive admittance (we're considering a parallel RLC) with the capacitive admittance.

This way, at resonance the LC parallel looks like an open circuit and the power from the generator only goes to the remaining R load in parallel.

At resonance, there is a current flowing alternately between L and C, satisfying from one side that

$$V = V_0 \, \sin(\omega_0 t)$$

across both the cap and the inductor (and the R), and on the other side that:

$$I_C = C \dfrac{d}{dt} \, V_0 \, \sin(\omega_0 t)$$

$$I_L = \dfrac{1}{L} \, \int^t V_0 \, \sin(\omega_0 t) dt$$

And you can develop the equations above to find out that

$$I_C = -I_L \, \, \, \, (\omega_0 = \dfrac{1}{\sqrt{LC}})$$

which respects everything about the single component and also their work together at resonance $\omega_0$ - In particular, the current in the cap peaks when the voltage across it goes through zero, and goes through zero when the voltage is approaching max.

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  • $\begingroup$ I think you are missing something here: (that Q) -> (that Q = 0) or (that Q = 0 at time 0) $\endgroup$
    – Frodyne
    Mar 27 at 13:46
  • $\begingroup$ Hmm, not exactly, I meant to say "that Q we're using in the V = Q/C formula". But I see from your comment that the phrase is confusing, so I'll try to fix it. Thanks! $\endgroup$
    – Mike Anblips
    Mar 27 at 14:15
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"Charge a capacitor" as it sounds, means that $\frac{\partial \rho}{\partial t} ≠ 0$

From the charge continuity equation this means that $\nabla \cdot \vec{J} ≠ 0$, and hence the current density $\vec{J}$ is non zero.

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It said that on the output side of transformer's secondary, when the voltage is higher, there's no current flowing because of highest impedance, and when voltage starts to reverse, the impedance starts going down and when voltage is zero, the current becomes highest.

It would be good to see the exact wording, because it sounds like you might be misunderstanding what was said. One way to visualize voltage, current, and charge in AC is to imagine a wheel whose axis is the origin of the Cartesian plane, and a red dot is marked on one point on the edge of the wheel. Now imagine the wheel rotating counter-clockwise. The y-coordinate of the red dot represent charge, and the x-coordinate represents current.

Now imagine putting a blue dot on the wheel whose x-coordinate represents voltage. The angle between these two dots represents the phase between current and voltage. If the dots are on the same point, then they are completely in phase, and power will flow from the source to the destination. If they're on opposite sides of the wheel, then they're completely out of phase, and the power will flow in the opposite direction. If they're ninety degrees from each other, then they are orthogonal to each other, and the direction of the flow of power will alternate back and forth, and no net power will be delivered in either direction.

What you are describing with "when voltage is zero, the current becomes highest" is this orthogonal case; when the blue dot is at the point (1,0), the voltage is zero, and the red dot is at the point (0,1), and so the current is maximum. It's theoretically possible for a circuit to have current and voltage be perfectly orthogonal, but it generally doesn't happen unless it's being deliberately sought.

What I suspect is being described is a situation where they are slightly out of phase, so that when the voltage is zero, the current is non-zero, but not entirely maximal. The average power delivered is zero in the orthogonal case, maximum in the perfectly in phase case, and in between when the voltage and current are partially out of phase. So the purpose of PFC is to bring the phase between the two to as close to zero as practical.

So voltage and current are out of phase, when voltage exists, current doesn't exist, and when current exists, voltage doesn't exist.

This implies that whenever voltage exists, current doesn't exist. But that's impossible. What is possible is that some of the times (rather than all of the times) that voltage exists, current doesn't exist.

So what my understanding of PF correction with caps is that, when load is connected to the output of a transformer's secondary along with cap in parallel, the capacitor gets charged when the voltage is higher and current is lower or zero, and when the voltage in secondary drops, the current from secondary goes directly to load because cap in parallel is already fully or 63% charged, at that time when voltage in secondary is zero, the charged capacitor delivers the voltage to the load (and why not back to secondary?) and this way the load receives the voltage and current at the same time and it consumes most of the available power.

So by my study, what I understand is for PF correction, the capacitor gets charged only wiht voltage and no current. However what I also studied is that there has to be current flowing in order to charge a capacitor.

First, as I explained before, there's generally going to be some current. You can't have a circuit where voltage and current and mutually exclusive. Second, there are different types of current. Any time a capacitor is charging, there is displacement current, which may correspond to an overall current, but it is different. If PFC is needed in the first place, then presumably the circuit has some inductance, and that inductance causes a displacement current. Ultimately, the purpose of the capacitor is to create a displacement current that cancels out the inductive displacement current.

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  • $\begingroup$ So you say that when the voltage is highest, the current is not zero, but minimum, and this minium current is enough to charge the capacitor that can induce enough voltage in load when current in secondary is highest and voltage is minimum (not zero), is that correct? $\endgroup$
    – Yogie
    Apr 1 at 12:49
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Charge is the integral of current over time. So there is no way of changing the charge of anything, including capacitors, without a current flowing.

Briefly, charge is the number of electrons in a given place, and current is the movement of electrons from one place to another. If you want to change the charge, you have to move electrons, i.e. a current must flow.

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