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Suppose in a home, power factor drops to zero. It means that the circuit becomes purely inductive.
If the energy meter is connected at incoming power supply point,will the energy meter still keep on increasing? If yes, then where is this energy being consumed?
We know that reactive power keeps on oscillating between source and load, so in this case, how does the reactive power fits into picture?
If there is GENUINELY ZERO resistive energy then energy will not dissipate, an ideal energy meter will not increment and the ideal reactive components will maintain the situation indefinitely.
Then, you wake up.
In the "real world" there are NO genuinely reactive components.
Inductors can resistance (and some capacitance).
Capacitors have ESR (resistance) and some inductance.
Wires have the same.
The energy meter shunt is often resistive, and even if it uses a Hall sensor it will not be absolutely lossless.
In the very very "best" case, it may take a wee while for things to die down. But never anything like indefinitely. You can work out the order of time for yourself by looking at eg published capacitor ESR values and inductor resistance, and wiring resistance. These set a lower level on losses before you start to look at the other (usually) second order components.
Please note there is a difference between power factor (PF) and displacement power factor (DPF)
PF is associated with the harmonic content while DPF is associated with the fundamental
DPF = cos(\$\Theta\$)
PF = \$\frac{DPF}{\sqrt{1+THD^2}}\$
why is this important? Because there are two scenarios where the real power can be zero.
1) Purely reactive load causing 90° phase shift between voltage and current
2) Purely harmonic current draw, causing no current draw at the fundemental.
colloquially, PF is used interchangeably with DPF more freely than it should be. In the scenario you are considering, the PF would equally the DPF as the THD = 0
What would happen? Well assuming you could acquire an ideal inductor and you had ideal cabling to your house, you would draw current equal to \$ \frac{V}{j\omega L} \$ while drawing no real power. This current could potentially be quite high.
Domestic billing is measured and charged in WATTS so you would theoretically not be charged anything. The grid and the generation however still need to source this current and this is where the problems are. Industry are charged in VA so every harmonic or DPF they draw, they must pay for since the generation and distribution must manage this.
If you or an industry complex did draw excessive DPF or PF there would eventually be a representative from the local electrical board to investigate (how quickly depends on how much disturbances since this impacts the generation, the distribution and everyone at the PCC)
In practice you cannot have an ideal inductor and thus there would be real power being pulled to feed the resistive element of the inductor and the cableing while also drawing (potentially) excessive current and DPF
Not only do all (non-superconducting) reactive devices have a non zero resistance, but also, any circuit of finite size (larger than a single electron) has a loop area, and will radiate EM energy. Both will heat up the universe, conserving all the added energy. Or something will melt.
In a reactive, oscillating circuit energy is dissipated by:
a) Resistance
b) Electromagnetic radiation
c) Gravitational radiation from the accelerating charges (negligible)
d) Changes in dimensions of components resulting in mechanical heat losses and sound
