You are under the misconception that you can control voltage and current independently of the resistance in the light bulb. Once you fix two of these three values, the third value is pre-determined.For example in a light bulb, if one keeps the current constant but increases the voltage the power is higher and hence more light generated. But why ? Why does voltage do this?
If for example one decreases the current as much as possible, to nano amps for example, but increases the voltage so high as to get the same amount of power as before, will the bulb shine just as bright ? Even tho current is so small ?
In transformers and long distance transmission lines, the current is so small, but the voltage is so high, yet when the signal arrives, its just a normal signal even though current was so low. But look, if the resistance of the line is high, which it is, then how do you recover a high current on the other side? Because the resistance of the line is so high, the current so low, so on the other end of course you will need higher currents for things, so how can you get a higher current from this high impedance line ?
Thanks everybody
Hello everyone, Firstly I'd like to welcome myself to ElectroTech.
I have a question about the role of voltage in power. Why is it that higher voltages generate more energy/power ?
For example in a light bulb, if one keeps the current constant but increases the voltage the power is higher and hence more light generated. But why ? Why does voltage do this?
If for example one decreases the current as much as possible, to nano amps for example, but increases the voltage so high as to get the same amount of power as before, will the bulb shine just as bright ? Even tho current is so small ?
In transformers and long distance transmission lines, the current is so small, but the voltage is so high, yet when the signal arrives, its just a normal signal even though current was so low. But look, if the resistance of the line is high, which it is, then how do you recover a high current on the other side? Because the resistance of the line is so high, the current so low, so on the other end of course you will need higher currents for things, so how can you get a higher current from this high impedance line ?
Thanks everybody
I am not very sure. If on the other end you receive the signal, but the impedance of the line is very high, how do you make that into a high current signal as needed? The line resistance limits any high currents. If you use a step down transformer, so that the voltage goes down, will the current automatically go high ?
I really dont get your electrons mixing thing. What do you mean by mixing electrons? A transformer doesnt mix electrons ..............
I am going to repeat that this is incorrect. If the light bulb has a fixed resistance you CANNOT keep the current constant while changing the voltage. If you increase the voltage, the current will also increase. The increase voltage AND current will cause increase in power.For example in a light bulb, if one keeps the current constant but increases the voltage the power is higher and hence more light generated. But why ? Why does voltage do this?
Hello everyone, Firstly I'd like to welcome myself to ElectroTech.
I have a question about the role of voltage in power. Why is it that higher voltages generate more energy/power ?
For example in a light bulb, if one keeps the current constant but increases the voltage the power is higher and hence more light generated. But why ? Why does voltage do this?
If for example one decreases the current as much as possible, to nano amps for example, but increases the voltage so high as to get the same amount of power as before, will the bulb shine just as bright ? Even tho current is so small ?
In transformers and long distance transmission lines, the current is so small, but the voltage is so high, yet when the signal arrives, its just a normal signal even though current was so low. But look, if the resistance of the line is high, which it is, then how do you recover a high current on the other side? Because the resistance of the line is so high, the current so low, so on the other end of course you will need higher currents for things, so how can you get a higher current from this high impedance line ?
Thanks everybody
I'm an EE.Have you got a degree in Elecronics or are you a student ?
I'm the electronics guy at a magnetic research lab.What do you work with friend ?
No. Nguyen is a Vietnamese, not that I've ever been there. I'm located in Canada.Are you from Hokkaido ?
As much as anyone else. When I say magnetics, I mean things like transformers, motors, and generators. I'm not the magnetics guy at the lab though.Do you understand about cassette tape and floppy disk drives?
Sorry, I don't Facebook people I don't know in person.So can you tell me about transformers please. Also I need your facebook so we can become friends. I like to be friends with EE people. I am also EE student .
So now, for a transformer, you apply a voltage on the left, and the voltage on the right will be Vs = Ns/Np * Vp, but will the current be precisely given by ohms law? So that Is = Vs/Rs ? or is there a catch ? Say the couplingficcient is 0.999999999999999999
Sorry, I don't Facebook people I don't know in person.
I assume p = primary and s = secondary.
No, the current will NOT be Is = Vs/Rs.
The current will be Is = Np/Ns * Ip.
This is because stepping up the voltage will step down the current, and stepping down the voltage will step up the current.
If there is a Resistor Rload across the secondary though, the current would be:
Is = Vs/Rload.
The reason is there's more than just Ohm's law inside a transformer. There's all the magnetic stuff happening. I'm not qualified to describe it all and even if I were, I would not be able to online.
Hello everyone, Firstly I'd like to welcome myself to ElectroTech.
I have a question about the role of voltage in power. Why is it that higher voltages generate more energy/power ?
Thanks everybody
Actually the current on the secondary is Vs/Rs......... Because the whole voltage appears across the transformer and hence resistor. So yes it is... Sorry
You were vague what your circuit actually was since you just posted bunch of subscripts without saying what they represented. Therefore, I assume an unknown load on the secondary and interpreted Rs as the internal resistance of the secondary winding (although it seems I forgot to make this clear in my answer, but I went back and added that line in) and gave an answer that would universally hold true:
Is = Np/Ns * Ip.
I then gave an additional valid solution if the unknown load was shown to be a resistor, Rload:
Is = Vs/Rload.
My answers already covered everything you said. The only new piece of information you gave was that you meant for Rs to be Rload. It doesn't change the correctness of my answers since I already covered all possibilities.
I actually did answer your question by talking about Rload separately, despite your spitefulness not allowing you to accept that. Where I am from, Rs is the winding resistance of the secondary and I had to make assumption because you assumed nomenclature was the same everywhere and did not bother to specify what you were actually referring to.You didn't cover all possibilities. You were wrong about the secondary resistance which is a load. You were incorrect sorry
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