• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

What is the role of voltage in Power?

Status
Not open for further replies.

AnalogueAlchemist

New Member
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
 

dknguyen

Well-Known Member
Most Helpful Member
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?
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.

Choosing a voltage and current means you are stuck with a single resistance value.

Choosing a current and resistance means there can only be one voltage value.

Choosing a voltage and resistance means there can only be one current value.

You can't say "I want this much current to flow through this much resistance at some arbitrary voltage."

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
Here is an analogy:

You can send me a tiny amount (low current) of very hot water (high voltage) down a narrow pipe (high resistance line) and I can mix it with a bunch of water on my end to produce a lot of water that is not quite as hot. Just like how I have water on my end, I also have electrons on my end. Electrons are everywhere. Loosely bound electrons are all inside the conductors on my end. The don't only have the electrons that I have aren't just the ones you are sending to me.

I can distribute the high energy in the small number of electrons that you send to me amongst many electrons on my end to produce a bunch of electrons that don't quite have as much energy. This is the same as how high voltage, low current can be sent down a line and turned into low voltage, high currents on my end.

In reality, electrons don't actually flow down the entire wire like water through a pipe. The electrons collide with each other and it's a chain reaction of collisions down the wire so the electrons don't really travel very far, very fast. It's just the energy from the collisions that travels. I can take one of your very energetic (high voltage) electrons and collide it with two of my low energy (low voltage) electrons to get twice as much current at half the voltage.

Does that help?
 
Last edited:

AnalogueAlchemist

New Member
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 ?
 

dknguyen

Well-Known Member
Most Helpful Member
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 ?
Did you follow my water analogy at all? The impedance of the line takes away from the voltage a little bit, but it REALLY takes away from the current. So you send high voltage at low current. Then on the receiving end, you mix the high voltage low, current with the electrons on the receiving end to make a higher current at lower voltage.

The step down transformer will automatically make the current higher when it steps down the voltage. Input voltage will always be higher than output voltage in a stepdown transformer. Likewise, input current will always be lower than output current in a stepdown transformer.
 

dknguyen

Well-Known Member
Most Helpful Member
I really dont get your electrons mixing thing. What do you mean by mixing electrons? A transformer doesnt mix electrons ..............
Is english not your first language? Let me know because it sounds like you're getting a bit hung up on certain words and analogies. I'm using analogies and the word mixing to try and simplify things but you seem to be missing the forest for the trees.

Mixing electrons is an analogy to mixing water. Mixing water at 1L of water at hot water with 1L of cold water to make 2L of warm water. With electrons, in actuality, you're not mixing but you are distributing the energy from a small number of high potential electrons across a larger number of zero potential electrons. That lets you turn high voltage, low current into low current, high voltage.

The answer to your question "how do you recover high current on the other side?" is that you distribute the energy from the received electrons amongst all the electrons already existing on the receiving side (in the metal conductors and such). The receiving side has lots of electrons, they just have zero potential energy until you give them a share of the energy being carried by the electrons being sent.

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?
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.

The only want to increase voltage but keep the constant current is to change to a light bulb with higher resistance. But this will not cause an increase in power. V = IR. Changing any two will determine the third. You can't manipulate all three independently of each other.
 
Last edited:

dknguyen

Well-Known Member
Most Helpful Member
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
Have you got a degree in Elecronics or are you a student ?
I'm an EE.
 

AnalogueAlchemist

New Member
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
 

dknguyen

Well-Known Member
Most Helpful Member
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, and Rs is the secondary winding resistance.

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.
 
Last edited:

AnalogueAlchemist

New Member
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.


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
 
Last edited:

Ratchit

Well-Known Member
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
Let's take one question at a time. First of all, voltages don't generate power. A dynamo or similar machine generates electrical energy at a rate called its power. A light bulb consumes electrical energy at a rate of power. Power is the rate of electrical energy consumption/generation with respect to time. It is measured in joules/second which are watts in the MKS system.

Voltage is the electrical energy density of a designated unit of charge. It is measured in volts, which is joules/coulomb in the MKS system. If you move a coulomb of charge through a one volt electrical energy density change, you will expend one joule of energy. If you do it in one second, you will expend energy at the rate of one watt of power.

Now, ask a question pertaining only to voltage, not transformers, until we get this point straighten out.

Ratch
 

dknguyen

Well-Known Member
Most Helpful Member
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.
 
Last edited:

AnalogueAlchemist

New Member
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.

You didn't cover all possibilities. You were wrong about the secondary resistance which is a load. You were incorrect sorry
 

dknguyen

Well-Known Member
Most Helpful Member
You didn't cover all possibilities. You were wrong about the secondary resistance which is a load. You were incorrect sorry
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.

But you will come to learn in your engineering education how to detailed you have to be when communicating with others. You are also going to need to grow a thicker skin if all it takes to be offended is for someone you have never met to not want to add you on Facebook.
 
Last edited:
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top