# Basic electronics theory (response to Layman's issues)

Discussion in 'Architecture & Engineering' started by billvon, Nov 18, 2014.

1. ### TrippyALEA IACTA ESTStaff Member

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Correct, and this is the exact same thing that myself, Billvon, Origin Leopold, and others have been saying to you.

A changing magnetic field creates a current, but a static one does not.

The duration of the emf mentioned only lasts as long as the change in magnetic field does.

No, you haven't understood it.

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3. ### LaymanTotally Internally ReflectedValued Senior Member

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If you refuse to believe that a magnet could create a voltage level in a circuit that is tied to ground, then you will never see the whole reason why I think the circuit would operate.
That is true. I agree with you there 100%. The part you don't realize is that it would create a voltage level. Then by adding components to that setup, you can create a current. Then without the proper setup, it would not create current.

I don't know how else to explain this to you. It would change the dipole moment of the wire. A power source does the same thing. It is just a place that has a certain charge. That charge can be transferred through the electromagnetic field. In other words, you could have a power source in series with a capacitor and it wouldn't make any difference. Just like it is common for capacitors to be in series with ground. In AC, the dipole moment of the power supply is just changing directions. If an electromagnetic field is applied to a line, it would change the dipole moment of the line in the same way as a power source. With a proper power source, the inductors would then function from that.

I distinctly remember running into some circuits when I studied electronics. Then I came across some circuits that seemed like it would be impossible for them to operate like they did. Then inductors can be used in the same way as capacitors that can step up and step down a voltage source. Then it is not very often that they are used that way, but essentially what you get is a capacitor that can step up the voltage. That only becomes possible if current is allowed to flow to somewhere else in the circuit.

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5. ### billvonValued Senior Member

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No, it won't. Again, voltage in a secondary loop is equal to frequency times the current in the first loop times the mutual inductance. When F is zero, V is zero. You can't get around physics no matter how many posts you make.
The opposite way, actually, because capacitors operate in just about the opposite manner that inductors do, due to the very basic difference in magnetic vs electric fields.
?? Not used that way very often? The PC you are using probably has both capacitive doublers (to generate the RS-232 levels in its serial port) and inductive boost circuits (to generate the voltage to excite the backlight tubes) in it. It's quite common.
Current can always flow "somewhere else in the circuit." Kirchoff's current law applies.

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7. ### leopoldValued Senior Member

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you might be confusing something layman.
electrostatic force, the effect when two electrically charged bodies approach one another.
magnetic force, that stuff that comes from magnets, an example would be lodestone.
the 2 are not the same, although they produce similar effects.
i guess this is one of the reasons einstien looked for a unified field theory.

8. ### LaymanTotally Internally ReflectedValued Senior Member

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Basically, a permanent magnet would be equivalent to having an inductor next to a dielectric plate. The inductor would create a charge in that plate.
That is not what I was saying. It makes it harder that I can show a diagram of the arrangement. If you had a circuit with an inductor next to an inductor of another circuit, then they would act capacitively to each other.
Most electronics are energy efficient. It would only be necessary to create extremely high voltages. In most electronics, just normal capacitors could be used to serve this type of operation.
Not if it has an open in it.

9. ### LaymanTotally Internally ReflectedValued Senior Member

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Basically, it is just that I have seen DC circuits that use inductors that then step up the voltage level to other inductors where the two inductors act like a really strong capacitor. I explained this possibility by re-describing Faraday's Law to the fact that it doesn't mention voltage level only current or power levels. Then some high power systems take advantage of this type of operation of an induction stepping up a voltage level to another inductor via the electromagnetic force in a similar way a capacitor would, but a capacitor just wouldn't step up the voltage.

10. ### billvonValued Senior Member

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Voltage = mutual inductance * frequency * current. You cannot get around that. You may believe you can, but physics disagrees - and physics always wins. From Wikipedia:
==========
Mutual inductance occurs when the change in current in one inductor induces a voltage in another nearby inductor.
==========
No, they would couple via mutual inductance. (That's how transformers work.) You might get some parasitic capacitive coupling but it would be quite small in comparison.
As in LCD screen backlight tubes.
Then current cannot flow. (And no, a capacitor is not an "open" even if you think the symbol looks like that.)
So - a transformer. Yes, they do exist.
Re-read it. It does indeed describe voltage.
Faraday's law never mentions power.
Again, they are used the opposite way capacitors are used. You really need some background in switching power supply design.

11. ### LaymanTotally Internally ReflectedValued Senior Member

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The troll always wins. That may be one way to calculate voltage but in no way is it required for there to be a frequency or current in order for a voltage level to be present.
Yes, they do behave the opposite way capacitors do when they are in series in a line.

12. ### LaymanTotally Internally ReflectedValued Senior Member

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A transformer can operate in much the same way as a capacitor, it is just that plates can be replaced by coils.

13. ### leopoldValued Senior Member

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they would react inductively.
they would transfer power electromagneticly

what are you talking about?

i'll scrounge around my HDD and see if i can come up with something for you.

14. ### billvonValued Senior Member

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It is the law of mutual induction. Again, there is no way for a static magnetic field to generate a voltage. Here's a basic primer on electronics theory:
=================
Electromagnetic Induction
We have seen previously that when a DC current pass through a long straight conductor a magnetising force, H and a static magnetic field, B is developed around the wire. If the wire is then wound into a coil, the magnetic field is greatly intensified producing a static magnetic field around itself forming the shape of a bar magnet giving a distinct North and South pole.

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Air-core Hollow Coil

The magnetic flux developed around the coil being proportional to the amount of current flowing in the coils windings as shown. If additional layers of wire are wound upon the same coil with the same current flowing through them, the static magnetic field strength would be increased.

Therefore, the Magnetic Field Strength of a coil is determined by the ampere turns of the coil. With more turns of wire within the coil, the greater the strength of the static magnetic field around it.

But what if we reversed this idea by disconnecting the electrical current from the coil and instead of a hollow core we placed a bar magnet inside the core of the coil of wire. By moving this bar magnet “in” and “out” of the coil a current would be induced into the coil by the physical movement of the magnetic flux inside it.

Likewise, if we kept the bar magnet stationary and moved the coil back and forth within the magnetic field an electric current would be induced in the coil. Then by either moving the wire or changing the magnetic field we can induce a voltage and current within the coil and this process is known as Electromagnetic Induction and is the basic principal of operation of transformers, motors and generators.

Electromagnetic Induction was first discovered way back in the 1830’s by Michael Faraday. Faraday noticed that when he moved a permanent magnet in and out of a coil or a single loop of wire it induced an ElectroMotive Force or emf, in other words a Voltage, and therefore a current was produced.

So what Michael Faraday discovered was a way of producing an electrical current in a circuit by using only the force of a magnetic field and not batteries. This then lead to a very important law linking electricity with magnetism, Faraday’s Law of Electromagnetic Induction. So how does this work?.

When the magnet shown below is moved “towards” the coil, the pointer or needle of the Galvanometer, which is basically a very sensitive centre zero’ed moving-coil ammeter, will deflect away from its centre position in one direction only. When the magnet stops moving and is held stationary with regards to the coil the needle of the galvanometer returns back to zero as there is no physical movement of the magnetic field.
=========================================

They always behave the opposite way. Capacitors try to keep voltage constant, and will vary current dramatically to attempt to do this. This is why capacitors make good filters and temporary storage devices. Inductors try to keep current constant, and will vary voltage dramatically to do this. This is why inductors make good ignition coils and flash triggers.
Not even close - and if you attempted to do this, you would fail miserably.

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15. ### TrippyALEA IACTA ESTStaff Member

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According to what you're espousing, resting a lead from a multmeter on a permanent magnet should be enough to induce a current, yes?

16. ### TrippyALEA IACTA ESTStaff Member

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Transformers and capacitors operate using very different principles. Did you ever look at that resonant circuit I showed you?

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18. ### LaymanTotally Internally ReflectedValued Senior Member

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Just in case a thousands words wasn't enough...

http://en.wikipedia.org/wiki/Capacitor_voltage_transformer

"In its most basic form, the device consists of three parts: two capacitors across which the transmission line signal is split, an inductive element to tune the device to the line frequency, and a voltage transformer to isolate and further step down the voltage for the metering devices or protective relay.

The tuning of the divider to the line frequency makes the overall division ratio less sensitive to changes in the burden of the connected metering or protection devices.[1] The device has at least four terminals: a terminal for connection to the high voltage signal, a ground terminal, and two secondary terminals which connect to the instrumentation or protective relay.

In practice, capacitor C1 is often constructed as a stack of smaller capacitors connected in series. This provides a large voltage drop across C1 and a relatively small voltage drop across C2. As the majority of the voltage drop is on C1, this reduces the isolation level of the voltage transformer. This makes CVTs more economical than the wound voltage transformers under high voltage (over 100kV), as the latter one requires more winding and materials."

19. ### LaymanTotally Internally ReflectedValued Senior Member

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Then according to all your crank crackpot pseudo-theories, the operation of a capacitor voltage transformer is impossible. The input voltage here is clearly DC. The output voltage is DC as well. Good luck working on or designing any high power systems without this one.

20. ### billvonValued Senior Member

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Yes, this is a common way to get a little isolated power from a high voltage line without having to have the massive number of turns you'd need on the primary without the divider. In practice it's used on kilovolt lines to operate relays and the like.

Now let's return to your previous statement: "A transformer can operate in much the same way as a capacitor, it is just that plates can be replaced by coils."

If you replaced the capacitors in that circuit with coils, it wouldn't work. The capacitors plus inductor form a resonant network and the transformer isolates the resulting AC waveform from the secondary circuit. Without the capacitors you do not have resonance. If you tried it the circuit would rapidly destroy itself, hopefully before it killed someone.

We can all be glad you no longer do work like this.

21. ### leopoldValued Senior Member

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read the following layman, and try to understand it.
it explains, in detail, what circuit analysis is all about:

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22. ### billvonValued Senior Member

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No, both input and output are AC. DC will not pass through a transformer.

I am starting to think you are really a troll. No one with any formal training could be this misinformed.

23. ### TrippyALEA IACTA ESTStaff Member

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So... Where, in the portion you quoted, it talks about the line frequency, and, where, on the link you provided it talks about frequency response, you think they're ferring to DC?