What is the meaning of voltage and how is it related to energy and current?

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The discussion centers around understanding the concept of voltage, with participants clarifying key electrical terms. Voltage is defined as the potential energy per unit charge, represented as joules per coulomb. It reflects how much energy each charge carries as it moves through a circuit. Coulombs measure the amount of electric charge, while amperes indicate the flow rate of that charge. An analogy is drawn comparing electrical flow to hot water in a radiator system, where the water represents current and the heat energy corresponds to voltage. Participants emphasize that voltage indicates the energy difference between two points in a circuit, and understanding this relationship is crucial for grasping electrical concepts. The conversation concludes with a sense of clarity on the topic, prompting the original poster to consider further questions.
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My first post here is going to be a question. Hopefully you guys can help clear this up.

I simply have been unable to grasp the concept of voltage.

I know how to calculate it and all but I can't grasp it. :bugeye:

I know that:

Couloumbs : the actual amount of energy

Amps : speed at which the current moves

Voltage : potential difference (?)

so what the heck does it mean when you say that a battery can generate 3V ??

Thanks a lot guys
 
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Eemsee2 said:
My first post here is going to be a question. Hopefully you guys can help clear this up.

I simply have been unable to grasp the concept of voltage.

I know how to calculate it and all but I can't grasp it. :bugeye:

I know that:

Couloumbs : the actual amount of energy

Oh-oh. You've got some unlearning to do. Coulombs are the amount of charge, not the amount of energy.

Amps : speed at which the current moves

This isn't quite right either, though it's pretty close. Amperes are a unit of current. Current is a measure of the rate at which charge flows, i.e.

1 ampere = 1 coulomb of charge / second
Voltage : potential difference (?)

Voltage = energy per unit charge

1 volt = 1 joule / coulomb
 
Voltage is the potential energy of a field per unit charge.

edit: I am blind, he already said that.
 
Voltage is the ratio of energy to charge(J/C). A good way of thinking of it is "how much energy is each little charge carrying around with it" as it flows.

As the current enters a lamp, it converts the "carried around" energy to heat and light, and so leaves the lamp carrying less energy than when it entered. So, if you put a voltmeter either side of a lamp, it will record a "Voltage" - ie a difference in energy either side of the lamp.

A useful analogy (for some) is the hot water flowing around a radiator system in a house. The water is the current - it just flows round and round and is conserved - whereas the 'Voltage' is the heat energy it carries around (per volume). The temp of the 'in' pipe is greater than the temp of the 'out' pipe - ie there is a 'difference' between them.
 
So, let's see if I got it.

Couloumbs measure how much charge there is (+ or -)

Amperes measure the speed at which it flows

Voltage is he amount of energy the each charge carries


so what is the difference between the number of couloumbs and volts. sorry guys


thanks
 
Adrian Baker said:
A useful analogy (for some) is the hot water flowing around a radiator system in a house. The water is the current - it just flows round and round and is conserved - whereas the 'Voltage' is the heat energy it carries around (per volume). The temp of the 'in' pipe is greater than the temp of the 'out' pipe - ie there is a 'difference' between them.

so would the couloumbs be the amount of water in the pipes?

:cry: :bugeye: :confused:
 
Any charge has a certain amount of potential energy because of its proximity to another charge - this is defined by a) the amount of charge of the two particles and b) the distance between the two charges. This can be complicated by the presence of a third, fourth, etc charge, but keeping it simple by considering only two particles of fixed charge, the amount of potential energy of one charge will remain constant if its distance from the other is fixed (it may still move, but as long as the radial distance does not change, the potential energy of the charge is constant). Therefore around a fixed charge there is a shell-like range of positions that the other may have wherein it has constant potential energy. This is the equipotential of the fixed charge. In this simplified two-charge system, if the distance of one charge from the other fixed charge changes, it moves to a different equipotential and so its potential energy will change.

Now, like I said, the amount of potential energy depends on the charge of both of the particles. However, if you have one particle q1 of fixed charge and you were to place another particle q2 somewhere near it, q2 will have a certain potential energy due to q1. If you double the charge of q2, you double the potential energy of q2. Therefore you can say that at a certain distance from, or on a certain equipotential of, a fixed charge, the amount of potential energy of any other charged particle PER CHARGE of that particle is fixed. This fixed value at each equipotential is the voltage of that equipotential. So at its most basic level, voltage is the amount of potential energy PER CHARGE of a particle that the particle will have at a given distance from a fixed charge.

However, values of potential energy are arbitrarily assigned - we choose infinite distance as have zero potential energy by convention only. We could equally choose zero to be metre, a kilometre, a millimetre... whatever. However, regardless of at what distance we choose to have zero potential energy, the change in potential energy as a charge moves from one equipotential to another is the same (i.e. you can tell how much potential energy has increased ot decreased as a charge moves a certain distance without knowing what the initial or final potential energies actually were). Hence we are generally interested in changes in potential energy, and therefore differences in voltage.
 
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Thanks a bunch guys, I'm pretty sure I get it now.

Now I got to think of my next question :-p
 
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