Electronics > Beginners
nobody talking about switching PS wasting power on input filter caps?
Ice-Tea:
--- Quote from: radiolistener on August 24, 2019, 08:03:39 pm ---
--- Quote from: Ice-Tea on August 24, 2019, 07:49:50 pm ---Now integrate the power over time.
--- End quote ---
it will be zero.
--- End quote ---
You understand this demonstrates no real effort from the generator side is required, right? You know, the point you've been denying for two pages?
--- Quote ---But you didn't answered on the question. Do you agree that AC source consumes power from the capacitor? Yes or not?
--- End quote ---
I'm not sure what you mean by 'consume', but if you mean that the generator gets the power back/consumes the power it delivered before then yes, that's what I've been saying all along.
soldar:
radiolistener, what happens in real life is not under question. It is well understood by engineers and engineers all agree.
Either you are totally wrong about it OR you do understand it and you are using vocabulary, language and explanations which are totally confusing and understood to be wrong. In either case the problem is with you, not with the rest of the world.
radiolistener:
--- Quote from: Ice-Tea on August 24, 2019, 08:14:16 pm ---I'm not sure what you mean by 'consume', but if you mean that the generator gets the power back/consumes the power it delivered before then yes, that's what I've been saying all along.
--- End quote ---
Yes, I mean the power (which previously is delivered to the capacitor) is consumed back by AC source.
So, we applied some amount of mechanical power to AC generator, it produce some amount of electrical power which is used to charge capacitor. Then capacitor applied this power back to AC generator and it was consumed by AC generator.
So what happens to this power next? Does it affects mechanical rotation of AC generator? Or what happens with this power?
Does this power just disappears? Does it slow down the rotor of AC generator? Does it speed up t he rotor of AC generator? ;)
radiolistener:
--- Quote from: soldar on August 24, 2019, 08:44:49 pm ---In either case the problem is with you, not with the rest of the world.
--- End quote ---
You're said that there is no energy loss due to capacitor. But if there is no energy loss, then you're invented perpetuum mobile. I don't believe in perpetuum mobile. And this is why I don't agree with you.
And the rest of the world also don't believe in perpetuum mobile. Therefore do not attribute your opinion to the whole world. That's just your opinion.
Zero999:
--- Quote from: radiolistener on August 24, 2019, 06:38:09 pm ---
--- Quote from: Zero999 on August 24, 2019, 06:12:54 pm ---If the power is zero, then why would more mechanical energy be required?
--- End quote ---
Because energy generated with AC generator rotation was accumulated by capactior and then returned back to AC generator. AC generator consumed returned energy to make rotor rotation more hard.
So, if you don't apply more energy, rotor will slow down and stop and AC voltage will drops down to zero. In order to keep AC voltage at 311 Vpk 50 Hz, you're needs to apply more power.
--- Quote from: Zero999 on August 24, 2019, 06:12:54 pm ---
There's a clear gap in your understanding. You haven't considered the fact that the alternator can also work as a motor.
--- End quote ---
I considered that. But it can works as a motor if you feed it with in-phase AC. Which is not the case for capacitor. Capacitor feeds AC generator with phase shifted AC.
If AC generator rotating clock wise, the capacitor will feed AC generator in such way so it will rotate counter clock wise.
--- End quote ---
No it won't, the current is leading the voltage, which will be driving the alternator in the same direction.
--- Quote ---Just look, if it works like you say, it will turns into perpetuum mobile, which is not possible. It doesn’t bother you?
--- End quote ---
In a real machine there are losses, so of course that won't happen. Adding a capacitor will increase the mechanical load on the generator, but not to the same degree as the equivalent resistive load will, which is what you think will happen. The increase in load will be minimal, inline with I2R.
If the alternator, capacitor and wiring had no losses, then it would run forever, but it wouldn't be creating energy from nowhere, just storing it. If a resistive load were connected, it would slow down and stop.
--- Quote from: radiolistener on August 24, 2019, 08:51:53 pm ---
--- Quote from: Ice-Tea on August 24, 2019, 08:14:16 pm ---I'm not sure what you mean by 'consume', but if you mean that the generator gets the power back/consumes the power it delivered before then yes, that's what I've been saying all along.
--- End quote ---
Yes, I mean the power (which previously is delivered to the capacitor) is consumed back by AC source.
So, we applied some amount of mechanical power to AC generator, it produce some amount of electrical power which is used to charge capacitor. Then capacitor applied this power back to AC generator and it was consumed by AC generator.
So what happens to this power next? Does it affects mechanical rotation of AC generator? Or what happens with this power?
--- End quote ---
Where does the power go when it's applied back to the AC generator?
To simplify things, suppose the alternator is being driven by a motor. We have an old fashioned motor generator set, which we could be using to change the voltage and frequency. The secondary side has a capacitor connected to it. The whole thing is cooled down to the point of superconductivity.
Where are the losses, other than the tiny amount of electromagnetic radiation caused by the currents flowing in the loops of cable?
How does connecting a capacitor which will draw 230mA of current cause the same amount of power dissipated as a resistor with the same impedance? The answer is it doesn't, the capacitor results in no extra losses, other than the tiny increase in electromagnetic radiation.
--- Quote from: radiolistener on August 24, 2019, 09:05:58 pm ---
--- Quote from: soldar on August 24, 2019, 08:44:49 pm ---In either case the problem is with you, not with the rest of the world.
--- End quote ---
You're said that there is no energy loss due to capacitor. But if there is no energy loss, then you're invented perpetuum mobile. I don't believe in perpetuum mobile. And this is why I don't agree with you.
And the rest of the world also don't believe in perpetuum mobile. Therefore do not attribute your opinion to the whole world. That's just your opinion.
--- End quote ---
No one has ever implied any perpetual motion.
You seem to be implying the reverse, i.e. energy can be destroyed. Energy doesn't simply disappear. It has to be converted into another form.
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