Question about Reactive Power and Displacement current

I've just starting studying about the Power Factor. As you know, there are 3 types of energy in the Power triangle: 1) Aparent power, 2) Active Power and 3) Reactive power.
The more the Reactive power, the less Active power used by the device.

My question is why reactive power is not useful to perform work?
I've read that Displacement current is a time-varying electric field that not has moving charges (electrons). If you don't have moving charges you don't have amperage, so you cannot perform work. Right?

And the last question: Reactive current is the same as Capacitive current?

J_Nichols,

I've just starting studying about the Power Factor. As you know, there are 3 types of energy in the Power triangle: 1) Aparent power, 2) Active Power and 3) Reactive power.

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There are two complex components of power in sinusoidal voltage/current driven system. The phasor sum of the two components is complex power. Apparent power is the magnitude of the complex power.

The more the Reactive power, the less Active power used by the device.

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That's wrong. The real or active power is independent of the reactive power.

My question is why reactive power is not useful to perform work?

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If reactive power were harnessed to perform work, then it would not be reactive poiwer any more. It would be real power. Reactive power is expended to build up the electromagnetic fields of capacitors and inductors, and returns back to the circuit when those fields collapse.

I've read that Displacement current is a time-varying electric field that not has moving charges (electrons). If you don't have moving charges you don't have amperage, so you cannot perform work. Right?

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That question in not related to the power triangle, so it should be in another subject thread. However the link below will give you more insight than you ever wanted to know.

https://www.electro-tech-online.com/threads/ac-flowing-through-a-cap-what-actually-happens.116575/

And the last question: Reactive current is the same as Capacitive current?

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Why are you capitalizing common adjectives? Reactive current/voltage/power covers both inductive and capacitive elements.

Ratch

The more the Reactive power, the less Active power used by the device.

Click to expand...

That's wrong. The real or active power is independent of the reactive power.

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Ok, but I've understood that when you've more reactive power in a system, the power factor decreases, and the system has more looses.
If the power factor is 1, there is minimal reactive power, so there is almost no looses.
Right?

If reactive power were harnessed to perform work, then it would not be reactive poiwer any more. It would be real power. Reactive power is expended to build up the electromagnetic fields of capacitors and inductors, and returns back to the circuit when those fields collapse.

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A friend of mine has built a circuit where he apparently puts in the primary of a transformer's core reactive current and he gets in the secondary real power.

That question in not related to the power triangle, so it should be in another subject thread. However the link below will give you more insight than you ever wanted to know.

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Well, maybe not, for that reason I made the question. But displacement current and reactive current cannot perform work, so I thought they would have something in common.

Why are you capitalizing common adjectives? Reactive current/voltage/power covers both inductive and capacitive elements.

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So, reading your answer I understand reactive current = capacitive current. That energy is measured in var. But, what this energy is composed of? Pure voltage without current like the output of some tesla coils?

J_Nichols,

Ok, but I've understood that when you've more reactive power in a system, the power factor decreases, and the system has more looses.
If the power factor is 1, there is minimal reactive power, so there is almost no looses.
Right?

Click to expand...

High power factors contribute to more transmission losses due to increased IR losses. That is because a higher current is needed to sustain the same power delivery to the load. That does not necessarily mean that less power will be delivered to the load. The apparent power can be increased to compensate for the increased IR losses. Your statement was that real and reactive power were linked. That is not necessarily true. Especially if the voltage can be increased, or if the current or resistance are low.

A friend of mine has built a circuit where he apparently puts in the primary of a transformer's core reactive current and he gets in the secondary real power.

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You will have to be more descriptive for me to understand what he is doing.

Well, maybe not, for that reason I made the question. But displacement current and reactive current cannot perform work, so I thought they would have something in common.

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They are two different things. Read the link.

So, reading your answer I understand reactive current = capacitive current. That energy is measured in var. But, what this energy is composed of? Pure voltage without current like the output of some tesla coils?

Click to expand...

Reactive power is the rate that energy is used to build and collapse electric fields for capacitors and magnetic fields for coils. It happens twice every cycle. In other words, energy is stored and released in capacitors and coils. The fields of capacitors and coils contain energy.

Ratch

A friend of mine has built a circuit where he apparently puts in the primary of a transformer's core reactive current and he gets in the secondary real power.

You will have to be more descriptive for me to understand what he is doing.

Click to expand...

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I'm reading the whole post, I will reply to the rest after.

He uses like a tesla's coil. In the primary he is pulsing DC, 15 Volts, 1,5 Amperes, at about 3Khz.
In the secondary he is using only 1 wire as the output. The other wire of the secondary is connected to the earth (to a metallic tube). This transformer is an air-core, step-up.

Using only one wire of the secondary he uses it as the primary coil of an step down transformer. In the secondary coil, output, he add an incasdescent light bulb and it lights. In this case, the core seems to be ferrite.

My questions are: Can be considered the energy flowing in the one-wire capacitive power? If the answer is positive, Why is possible to perform work with capacitive power?
Sorry if I've made some mistake wrtitting the concepts.

Hi,

He found a secret vortex beam from another galaxy seeping into our galaxy and nobody knew it until now.

Seriously though, when you do things like this you have to make very careful and precise measurements of the entire system, input and output, and find out where the losses are. The output never shows as much power as the input, as some is always lost as heat in the system somewhere.
To say that he is somehow using "capacitive power" doesnt mean too much because a pure capacitance doesnt absorb any power. So you'd have to explain that better and probably take some measurements including the phase of both input and output.

It would help enormously if you would draw up a schematic with the elements shown as they really are, and what you think they are doing. That way we can either find out where the losses are or finally be able to tap into that secret vortex

(Delete)

Oh yes!, he found a vortex and then he was absorbed by it. For that reason I haven't see him since a lot of time.

Well, coming back to the question. The only thing I have is this video

I would like to give all those data to you but I've no the things to perform the test by myself. The thing I cannot understand is that the energy input to the primary coil of the transformer comes from a single-wire transmission line. So, I've no idea what kind of energy is flowing there. Electromagnetic or static? The output is rated at thousands of volts and I can touch the cable and I don't get shocked. So I think the energy is pure static without amperage, because high voltage and amperage will kill anyone.

So you didnt get killed so that's how you know it works?

No, I know it works because the bulb lights. I thought if I didn't get shocked with high voltage then it should be zero or near zero current.

But comingn back to the question. What kind of energy flows in the single-line transmission wire? because if I connect to a device only one pole of a battery it doesn't works.

Hi,

What kind of bulb is that?