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Farmhand

• Hero Member
• Posts: 1583
Re: Silly question about voltage and current
« Reply #30 on: February 27, 2014, 12:28:24 AM »
I have a question: Do the voltage and the current for a cetrain load have to come from the same source?

Or could I, example given, use two independent sources like photopholtaic and wind or whatever, then go out of phase with each one by -90 resp. +90 degrees with a cap and a coil, then ac- sync this so in none of them is actally both (v+a) present, then simply but these together , maybe trough some diodes, to get a complete, working source of V+A ?

It works like this as I see it. If the two supplies are electrically isolated until the outputs are merged, then each circuit will take it's own return current only.

ie, The HV circuit supply voltage causes the HV circuit's current and so the current caused by the HV will return to the HV supply.
The LV supply circuit's voltage causes the low voltage circuits current and that current will return to the LV supply.
Now if the HV and LV output are simply superimposed and have common grounds the HV will see the LV as a load.  Like connecting the output of a flyback transformer to a battery. LV is load to HV.
If the reverse current is blocked to prevent that then the load determines what gets loaded the most.
If the load is high resistance the LV supply will do no work.

It's really quite simple and comes down to what pushes what, and current loops in separate supply's. It's not rocket surgery.

The voltage causes the current to flow in a (loop). Each supply will have it's own loop.

If we talk AC and we apply 10 000 volts positive half cycle to 150 volts negative half cycle, there will be a potential difference of 10150 volts, if the two circuit grounds are common the HV supply will see the LV as a load.

Cheers

EDIT: I talk in 'conventional' positive to negative current, not "electron current'. Flow of 'charge' (or whatever), not electron movement. Just to say how I think with respect to "current" flow.

It should be noted and logic that if the two were superimposed on the same conductors then the HV and the larger current will likely be measured, but they won't perform work together, the HV and it's associated current will do most of the work if the load is high enough resistance.

It basically conforms to work done at input and energy output. V x A only applies to the A caused by the V that caused it, and means power not energy anyway..

Spock and Kirk (Logic), Make up your own minds who is Spock and who is Kirk.
..

If we add a Flea to an Elephant will we end up with a Flea the size of an elephant ( would be a nasty animal), or an elephant the size of a flea, or will we just end up with, a Flea and an Elephant ?

..

Farmhand

• Hero Member
• Posts: 1583
Re: Silly question about voltage and current
« Reply #31 on: February 27, 2014, 01:16:16 AM »
This is kinda interesting and I think discussing it and thinking about it would help the less experienced to better visualize voltage and current.

IF we could post some drawings of situations where the mixing of HV-LC and LV-HC may be achieved then the circuits could be evaluated. And the effects seen in a picture.

The drawings would need both HV and LV supplies included with any groundings and the load/loads, supply - load isolation if any, ect. ect..

Pencil paper and eraser are cheap and work like a simulator to a degree, we just need to add the visualization based on what we know happens, though it can get complicated, it's often simple when seen on paper. I draw the supply itself then the circuit from positive to negative as much as is possible.

..

MarkE

• Hero Member
• Posts: 6830
Re: Silly question about voltage and current
« Reply #32 on: February 27, 2014, 01:35:33 AM »
This is kinda interesting and I think discussing it and thinking about it would help the less experienced to better visualize voltage and current.

IF we could post some drawings of situations where the mixing of HV-LC and LV-HC may be achieved then the circuits could be evaluated. And the effects seen in a picture.

The drawings would need both HV and LV supplies included with any groundings and the load/loads, supply load isolation ect. ect..

..
The relationships are not complicated.  Assign whatever voltage you like to each of the two sources.  Assign the corresponding internal resistance values.  The problem that you will run into is where the current limit of the high voltage supply causes the value of the associated resistor to be variable, and the resulting external voltage across the combination of Vx and Rvx to collapse.

Farmhand

• Hero Member
• Posts: 1583
Re: Silly question about voltage and current
« Reply #33 on: February 27, 2014, 01:39:16 AM »
But Mark those supplies are in series. I'm not sure that is what is meant by the original poster.

..

MarkE

• Hero Member
• Posts: 6830
Re: Silly question about voltage and current
« Reply #34 on: February 27, 2014, 01:45:11 AM »
But Mark those supplies are in series. I'm not sure that is what is meant by the original poster.

..
If you want the voltage of two supplies then you have to put them in series, and the loop current will be limited by the supply with the lower current limit.  A 25kV 1mA supply in series with a 2V 150A welder will still only deliver 1mA.  If you want the current of two supplies then you connect them in parallel, and you will be limited by the lower voltage of the two.

Farmhand

• Hero Member
• Posts: 1583
Re: Silly question about voltage and current
« Reply #35 on: February 27, 2014, 02:22:06 AM »
If you want the voltage of two supplies then you have to put them in series, and the loop current will be limited by the supply with the lower current limit.  A 25kV 1mA supply in series with a 2V 150A welder will still only deliver 1mA.  If you want the current of two supplies then you connect them in parallel, and you will be limited by the lower voltage of the two.

Umm you don't need to tell me that.  But is that what the original poster wants or is talking about ?

I think he is talking more about a HV in parallel with a LV maybe the HV as HF as well, I'm not sure. I'm hoping the original poster will post a drawing of what he is visualizing.

This below is what I think they mean when they talk of mixing HV with LV. Don't forget, it is no me making the claim here. First I am trying to understand what the original poster actually means.

Maybe he will say if my drawing represents his thoughts. It could be considered without the smoothing capacitor or with.

In my opinion it's a pointless exercise. Unless there is some reason for it that I cannot see.

..

Farmhand

• Hero Member
• Posts: 1583
Re: Silly question about voltage and current
« Reply #36 on: February 27, 2014, 02:44:49 AM »
Here's an example, If we begin with the LV supply turned off and the Load resistor drawn near the ? on the drawing is 10000 Ohms then the capacitor will charge to about 1000 volts and about 0.1 Amps will flow in the load resistor. about 100 Watts. Then if we turn on the LV supply nothing will happen, because the capacitor is already charged to 1000 volts. The LV supply will draw idle power. That's what I say.

Now if we do it the other way and have the LV supply powering a 100 ohm load and dissipating about 1 Watt. Then we turn on the HV supply we could have a problem with fire or supply capabilities, or just use more power (10 kilowatts) and drive the load harder with the HV supply.

Cheers

MarkE

• Hero Member
• Posts: 6830
Re: Silly question about voltage and current
« Reply #37 on: February 27, 2014, 02:53:44 AM »
Umm you don't need to tell me that.  But is that what the original poster wants or is talking about ?
Those are the only two options available:  Series or parallel.
Quote

I think he is talking more about a HV in parallel with a LV maybe the HV as HF as well, I'm not sure. I'm hoping the original poster will post a drawing of what he is visualizing.

This below is what I think they mean when they talk of mixing HV with LV. Don't forget, it is no me making the claim here. First I am trying to understand what the original poster actually means.

Maybe he will say if my drawing represents his thoughts. It could be considered without the smoothing capacitor or with.

In my opinion it's a pointless exercise. Unless there is some reason for it that I cannot see.

..
I think that the person proposing the scheme does not understand that all practical voltage sources have internal impedances that limit their current.

Farmhand

• Hero Member
• Posts: 1583
Re: Silly question about voltage and current
« Reply #38 on: February 27, 2014, 10:06:25 AM »
Mark ,Even to a novice it should make no sense to connect a thin wire from say a flyback secondary in series with the thick wires from a step down transformer secondary.

It is true the impedance of the supplies will see the HV supply restricted in output current and at some point the voltage will drop. But for the sake of the argument we can consider both transformers capable of outputting say 1000 watts each or we could consider them ideal and thus they would handle any load, just to keep it simple.

We can see that in the first case of the LV off to start with then turning the LV on while HV is already on and the load resistance is high makes little difference due to the counter emf of the already charged to 1000 volts capacitor would prevent any power output from the LV supply.

I think if we want people to understand more we need to give examples and explanations in layman's terms, eg, in my example the counter emf situation shows up another myth which is that counter emf (the evil twin or the evil witch as UFO calls it)  consumes power, when in fact it only restricts the power, counter emf does not consume power and any emf countering an applied emf is counter emf, be it induced or already present.

If we just say no it won't work and give technical explanations that most people do not understand then next to nothing is achieved. People don't just trust the words of a "naysayer" such as we, like they trust the words of the guru scammers. We must work to get the point across.

I am always amazed when a trained guy comes along and lumps me in with the claimants of OU or suggests I am making claims, when all I am doing is trying to explain how I see things in layman's terms.

I suggest if you want a strictly engineering interaction then go talk to an engineer. No offence but you cannot expect to talk engineer speak to laymen or complete novices and have them just accept it. If it bothers you not that the same fallacies keep arising then keep not explaining it so people can understand. This mixing of HV-LC and LV-HC has been claimed before by others and i challenged them to show it but was personally abused.

You should be aware Mark that I am on the side of (proof of claims and better understanding of basic electronics) as a means to reduce the effect of the Guru scammer type people and the eternal promise with no evidence - keep a thread going for ever type people, as well as helping others to be more discerning experimenters able to spot a fake or a mistaken claim.

Cheers

TinselKoala

• Hero Member
• Posts: 13968
Re: Silly question about voltage and current
« Reply #39 on: February 27, 2014, 02:00:08 PM »
The system in the RM503 scope is something like that shown below. I haven't put in the capacitors, etc. and I've shown it as a positive elevation instead of the negative "elevation" used in the real scope. The filament is elevated to -3000 v wrt the scope's chassis ground; this supplies the electron current needed to make the electron beam. The filament gets its AC heating current from the 12 v transformer, which can supply its full rated output to a load, and the filament gets the DC beam current from the HV transformer (really a voltage multiplier system) through the HV rectifier tube and this can only supply a small current.

Needless to say, the insulation in the 12v transformer has to be good.

The loops described by Farmhand are here; they share one "leg" in common. If you could be an electron in that common leg... you'd probably get pretty confused. But it works, anyhow.

dieter

• Hero Member
• Posts: 938
Re: Silly question about voltage and current
« Reply #40 on: February 27, 2014, 02:02:39 PM »
What I was talking about was: in AC a cap as well as a coil will bring the current out of phase. There is a moment in the AC cycle where there is full current at the cap, but zero voltage, and there's a moment where there's full voltage at the coil, but zero current. If you would use a mechanical or electronic switch to tap that very moment only and then "mix" the currentless voltage with the voltageless current, would anything happen?

But I do assume, even if the current is out of phase, energy is consumed nonetheless.

TinselKoala

• Hero Member
• Posts: 13968
Re: Silly question about voltage and current
« Reply #41 on: February 27, 2014, 05:12:49 PM »
What I was talking about was: in AC a cap as well as a coil will bring the current out of phase. There is a moment in the AC cycle where there is full current at the cap, but zero voltage, and there's a moment where there's full voltage at the coil, but zero current. If you would use a mechanical or electronic switch to tap that very moment only and then "mix" the currentless voltage with the voltageless current, would anything happen?
Uh.... no. Or maybe yes. You see, when you set up impossible initial conditions, it's impossible to predict outcomes. Sort of like dividing by zero.
Quote
But I do assume, even if the current is out of phase, energy is consumed nonetheless.
Not necessarily. As we covered before, AC power fed to reactive circuit components will exhibit phase shift between voltage and current. This results in the "power" being partitioned into "real" power and "reactive" power. The reactive power isn't dissipated.

From Wiki:
Quote
In the diagram, P is the real power, Q is the reactive power (in this case positive), S is the complex power and the length of S is the apparent power. Reactive power does not do any work, so it is represented as the imaginary axis of the vector diagram. Real power does do work, so it is the real axis.
The unit for all forms of power is the watt (symbol: W), but this unit is generally reserved for real power. Apparent power is conventionally expressed in volt-amperes (VA) since it is the product of rms voltage and rms current. The unit for reactive power is expressed as var, which stands for volt-ampere reactive. Since reactive power transfers no net energy to the load, it is sometimes called "wattless" power. It does, however, serve an important function in electrical grids and its lack has been cited as a significant factor in the Northeast Blackout of 2003.[2]
Understanding the relationship among these three quantities lies at the heart of understanding power engineering. The mathematical relationship among them can be represented by vectors or expressed using complex numbers, S = P + jQ (where j is the imaginary unit).
http://en.wikipedia.org/wiki/AC_power

Dave45

• Guest
Re: Silly question about voltage and current
« Reply #42 on: February 28, 2014, 01:04:26 AM »
So if voltage is pressure and current is flow tell me how there can be a phase shift between the two,

MarkE

• Hero Member
• Posts: 6830
Re: Silly question about voltage and current
« Reply #43 on: February 28, 2014, 02:03:21 AM »
So if voltage is pressure and current is flow tell me how there can be a phase shift between the two,
That's easy, set up a channel with an accumulator attached.  If you increase the pressure in the channel, more water will store in the accumulator, and the flow downstream will not increase in lock step with the pressure increase.  Likewise, when you decrease the upstream pressure, the accumulator will release water and the flow downstream will not decrease in lock step with the pressure decrease.  This is why municipal water systems have water tower accumulators.

Dave45

• Guest
Re: Silly question about voltage and current
« Reply #44 on: February 28, 2014, 03:06:52 AM »
Ok sounds plausible for water, tell me if alternating current is just current alternating through a conductor then why does this circuit build a negative ion cloud on one end and a positive ion cloud on the other.
If negative current were the only energy moving in the coil wouldnt we have a negative ion cloud only on one end of the circuit.