Capacitor and inductor power factor

Question

I need an explanation about the leading and lagging power factor at electron level.

Case1:
In a inductor it has the lot of free electrons are ready to move when we applied a voltage(AC or DC). If we apply a alternating voltage to a inductor, a change in current with respect to time is passing through it. According to induction principle, a change in current produce a change in magnetic field. So that a change in magnetic field produce a induced emf. This emf opposes the change in current wich was supplied by a alternating potential and which was causes of that emf.

Here My first question is for inductor: How the electron flow lags behind the supply voltage?

Case2:
For a capacitor there is a dielectric medium between two plates. The dielectric medium itself has the randomly oriented dipoles. So that the net electrostatic force is zero. When we apply a potential between these two plates, the potential start to align the randomly oriented dipole's positive sides towards in which plate the electrons are accumulated and the negative sides of the dipole towards in which plate the electrons are pulled away.

Here My question for capacitor is: How the electron flow (current) leads the voltage?
(according to electrical laws potential only causes of electron flow. So without a potential the electron flow doesn't possible. How the electron flow leads the voltage in capacitor?)

Answer 1

For an inductor, how does the electron flow (current) lag behind the supply voltage?

In an inductor which has no magnetic field already established, the application of electricity initially sees an open circuit (little current flow.) As the magnetic field builds, the current also builds, and it eventually behaves like a solid conductor (full current flow.) So the current lags the supply voltage due to the delay in establishing the magnetic field.

(If you then instantly disconnect the inductor, the magnetic field will collapse in reverse as quickly as it can. With no resistance to slow it down, dv/dt dictates that the voltage will go exponentially high. This is called "inductive kickback" and can be problematic, even dangerous.)

For a capacitor, how does the electron flow (current) lead the voltage?

In a discharged capacitor, the potential (electric field) between the two plates is nothing, so no current flows. Application of any electricity initially sees a short circuit (full current flow.) As the plates start charging up, the current decreases, and eventually behaves like an open circuit (little current flow.) So the current leads the supply voltage due to the delay in establishing the electric field.

(If you then instantly disconnect the capacitor, the electric field remains static, like static electricity. It may remain there for years, ready to zap you, such as in older televisions and radios.)

The key similarity between the two is that it takes time for magnetic fields to build/collapse and plates to charge/discharge; this delay creates an imbalance between the voltage and current measured at each device... and we call the ratio of this "Power Factor."

Answer 2

Let's assume a sinusoidal voltage source.

Case1:

At the start of the AC cycle there is a rapidly increasing voltage. This tries to push current through the coil. As the current rapidly increaces this creates a changing magnetic field. This creates an emf which opposes the very change in current (as you rightly said). This emf pushes electrons back in the opposite direction. This is what essentially leads to the lagging current.

1. initially the change is voltage is maximum, this creates a large back emf and hence a large current is "pussed back in the opposite direction"

2. As the voltage approaches its maximum the change in voltage is minimal and hence the back emf is minimal, and hence this backward current ceases to flow.

3. Now the change in emf becomes negative and hence it pushes the electrons in the forward direction. Hence forwards currents start to flow

If you plot this you will see that a sinusoidal input leads to a lagging output because the voltage is proportional to the change in current.

$$v_L=L\frac{di}{dt}$$

Case 2:

Similarly for a capacitor the current is proportional to the change in voltage. Therefore initially when the change in voltage is maximum the current passed around the circuit will be maximum. Now since the plates are saturated with charge adding more becomes difficult and the current flow decreases. As the voltage becomes maximum the current through the circuit becomes zero as the change in voltage is also zero.

$$i_c=C\frac{dv}{dt}$$

Answer 3

The electrons (or current) do not physically "lag" the voltage (in an inductor). This is because the voltage is proportional to derivative of current: $$ U = -\frac{di}{dt} \times L $$

Please consider DC: the current does not change, so there is no voltage drop on the inductor terminals. If you have current that changes, but not in sinusoidal way, the voltage drop is proportional to the current. For example, if current applies to the formula \$ i(t) = 5 + 2 \times t \$ (in Amps), the voltage drop would be constant, 2 Volts (even if the current increases to infinity), multiplied by L.

The AC is just a special case which takes into account that a derivative of sin function is the cos function and vice versa (with or without the minus sign). If the current would be \$ i(t) = 5 + 2 \times \sin(\omega\times t) \$ the voltage would be \$ v(t) = -L \times \frac{di}{dt} = -L \times 2 \times \omega \times \cos(\omega \times t) \$.

Because both voltage and current have the same frequency (\$ \omega \$), we can compare them (for example using complex numbers) and it looks like the current is behind the voltage (ie. it is lagging), but it only applies to ac.

In my opinion, when you use this method, it is not possible to understand the phenomenon on the electron level in other way you did. The complex numbers do not exists in real world, it is just a mathematical method of calculation, correct for this particular case only (ie. for sinusoidal alternating current). You explained correctly that currents makes the magnetic flux which then causes a voltage drop and this is all to be understood.

If you take any changing current and represent it with a Taylor series (see Fourier transform, Fourier series and relative articles), you may perform the calculations for any harmonic (\$ n \times \omega \$, with n being consecutive integers), on its complex plane, however you can't merge planes for different harmonics.

The Fourier method also is a mathematical tool, but of course the inductor does not "know" it should calculate harmonics for every signal it reads.

For capacitor the same applies.

(I don't know how to insert formulas, could anybody please edit my text?)

Answer 4

Partial answer regarding the capacitor, maybe later I can add the inductor.

Initially the capacitor is 'empty': the dielectric medium's dipoles are randomly (un)aligned. Now if you apply a sinusoidally alternating voltage to the plates, the plates need to be populated by electrons (and on the opposite side a lack thereof?).

In order to populate the plates with charge, per definition a current needs to flow. That is basically the transport of the charged particles to/from the plates.

If you have a capacitor with capacity \$C\$, then for every volt across the plate you need to populate with \$Q=CV\$ charge. For a constantly increasing voltage, lets say \$\frac{dU_C}{dt}\$, you constantly need to increase the amount of charge on plates in order for the voltage to rise constantly. For a rise in voltage \$\frac{dU_C}{dt}\$ you need a flow of \$C\frac{dU_C}{dt}\$ of charge to the plates. This is a DC current, say, \$I_C\$.

Now, if you apply a changing, but not constantly increasing, voltage to the plates, the flow of charge will obey the same law: \$I_C(t)=C\frac{dU_C}{dt}\$. Hence when you apply a sinusoidally alternating voltage \$U_C(t)=U.sin(\omega t)\$, the current to the capacitor will be \$I_C(t)=CU\frac{dsin(\omega t)}{dt} = \omega CUcos(\omega t) \$, which is leading the voltage. This is understandable as the rise in voltage is the greatest at the start of the sinus, and thus the current must be the biggest at that moment. While at the top of the sinus the voltage doesn't change and therefore no charge is supplied to or removed from the capacitor; therefore the current then is zero (for a moment).