Circuit can supply negative voltages ?

[Femme_physics]

Circuit can supply "negative voltages"?

I read an article trying to understand voltage dividers better.

I then read this quote that threw me off!

As you know, some circuits are designed to supply both positive and negative voltages. Perhaps now you wonder if a negative voltage has any less potential than a positive voltage. The answer is that 100 volts is 100 volts. Whether it is negative or positive does not affect the feeling you get when you are shocked.

Now, what does it mean that a circuit is designed to supply both positive and negative voltages? I'm pretty new to electricity. All I need is that a difference in potentials (i.e. voltage difference) means current flow. That would be conventionally from plus to minus (although electrons are considered to carry a negative charge). For all I know, that is the only possible type of motion. Negative electrons flowing (conventionally) from positive to negative.

Can anyone help clear this out?


The article's link:

http://www.tpub.com/neets/book1/chapter3/1-35.htm

 

[I like Serena]

A battery maintains as you say a "voltage difference".

This means the plus pole has a certain voltage and the minus pole has a certain voltage.
The difference between the two is the "voltage difference"

However, the height of this voltage is undefined.

To be able to get predictable absolute voltage amounts, we need to introduce a reference point.
The reference point that is usually used, is the "earth". (Note the special symbol for it.)
For that purpose a circuit is usually attached to "earth" to define "zero volts".

Often the minus pole is attached to earth, meaning that all voltages will be positive.

But alternatively, as in the diagram you referred to, one can attach the Earth between a couple of resistors.
The result is that any points closer to the minus pole will have a negative voltage relative to the reference point (earth).

EDIT: as for your quote. You get "shocked" by a "voltage difference", meaning one part of your body must make a connection with a different voltage than another part of your body.
If you use different hands to touch different poles, the current will go through your heart with possibly fatal consequences.

 

[Studiot]

I'm sure you have no difficulty with the idea of sea level, mountain height and ocean depth.

Positive and negative voltage works just the same; it's all about datum (reference level).

We can say that the mountain top is +1500 metres and the ocean floor -1500 metres, relative to sea level being zero.

The difference in elevation between the ocean floor and the mountain top is 3000 metres.

In electricity if we take a zero reference then we can arrange for some other points to be +15 volts or -15 volts relative to this point. Often we Earth or ground the zero point, but not always.

So if we take two 15 volt batteries and link the negative of one to the positive of the other and call this link zero then the second terminal of one battery is at +15volts and the second terminal of the other is at -15volts and the total difference is 30 volts.

go well

 

[KingNothing]

The numbers are all relative. A power supply that has outputs of -10V, 0V, and +10V is exactly the same as one that has outputs of 0V, +10V, and +20V.

Electrical engineers choose a point in a system to call 0V, or ground. Typically we like to choose the Earth itself as our "ground", or some other convenient part of circuit. For devices that get connected to our AC mains, we like to choose the ground plug of the power cord to be our 0V point.

It's all based on convenience.

 

[Femme_physics]

Well, if this is all about grounding points, how come we never see a "grounding" points in some problems? Why is it always "suddenly introduced"?

i.e.

Say I have a simple circuit

Say I know R.

Now I go about doing my calculations, using ohms law to solve for I. So great. Now I solved for current. Great. I solved the exercise without a grounding point. It's not written in the problem statement, not graphed, nothing! How is that possible? Was there really no grounding point defined? Or is it default? Or was it pre-defined to me when I was given the voltage? What's going on here?

Am confused!

 

[I like Serena]

Well, if this is all about grounding points, how come we never see a "grounding" points in some problems? Why is it always "suddenly introduced"?

i.e.

Say I have a simple circuit

Say I know R.

Now I go about doing my calculations, using ohms law to solve for I. So great. Now I solved for current. Great. I solved the exercise without a grounding point. It's not written in the problem statement, not graphed, nothing! How is that possible? Was there really no grounding point defined? Or is it default? Or was it pre-defined to me when I was given the voltage? What's going on here?

Am confused!

In your particular example it does not matter.
With or without grounding point, the result is the same.

In practice you always want to ground your circuits for a variety of reasons.

And even that is not always straight forward, since you want to connect to "only one ground", otherwise you'd get a so called "earth loop" which manifests as a loud hum in acoustic electronics.

@Studiot & KN: good replies. :)

 

[Femme_physics]

In your particular example it does not matter.
With or without grounding point, the result is the same.

Then in what types of problems does it matter?

I have two points in this problem if I got it right. So, now matter which point I pick to be my ground point, the result is the same? Hmmm... I think that's because I have only 1 resistor. What if I had 2?

 

[I like Serena]

Then in what types of problems does it matter?

I think that in none of the problems you will be given will it matter.

It usually only matters when applying electronics in practice.

Such as when you want to make an accurate measurement using an electronics circuit.
It will need to be properly grounded, or you'll see your measurements "drift".
(Although they may drift anyway.)

Or when there's only one wire being used to connect everything, such as in a car.
In a car, the plus pole is wired to for instance the radio.
But the other side of the radio is wired to the chassis (that is, the ground).
The chassis works as a "common return" wire.

I have two points in this problem if I got it right. So, now matter which point I pick to be my ground point, the result is the same? Hmmm... I think that's because I have only 1 resistor. What if I had 2?

That still won't matter.

In your examples you can "choose" your reference point.
If it bothers you, it's best to "choose" the minus pole to be zero volts, just as if it was grounded.

 

[Femme_physics]

I think that in none of the problems you will be given will it matter.

It usually only matters when applying electronics in practice.

But don't you have to understand it before you apply it? I'm not sure I do at the moment!

That still won't matter.
In your examples you can "choose" your reference point.
If it bothers you, it's best to "choose" the minus pole to be zero volts, just as if it was grounded.

And if I choose my + to be the zero point, will it make my voltage negative?

 

[I like Serena]

But don't you have to understand it before you apply it? I'm not sure I do at the moment!

What don't you understand?

As I said, it sets a reference point.
In theoretical problems it won't matter.
In practical problems it eliminates unwanted side effects.

And if I choose my + to be the zero point, will it make my voltage negative?

Yes! :)

 

[Femme_physics]

So what we weren't told that in the problem such as the one I posted, there IS in fact a ground reference point. V minus, like you said, is the 0 point! I wish it was mentioned to us, it'd make things a lot less confusing if we ALWAYS knew where the reference point is, even in simple problems :)

What don't you understand?

That cleared it, sugar :) Thanks!

 

[Femme_physics]

Just to run home with this:

So in this simple circuit, I chose to make my grounding point between R1 and R2.

Does it mean that my calculations settings IR1= -4 and IR2= -8 are correct? (the first image's calculations is without a grounding point. The second is)

http://img24.imageshack.us/img24/5607/cir1.jpg

http://img651.imageshack.us/img651/4791/cir2.jpg

 

[I like Serena]

Just to run home with this:

So in this simple circuit, I chose to make my grounding point between R1 and R2.

Does it mean that my calculations settings IR1= -4 and IR2= -8 are correct? (the first image's calculations is without a grounding point. The second is)

Sorry, no.
You're mixing voltages and "voltage differences".
And anyway, setting a grounding point won't make the current suddenly go the other way.

Now that you've set a grounding point, we can talk about "absolute" voltages.
The voltage of the plus pole will be 4 V, so the voltage "across" R1 is (4 V - 0 V) = 4 V.

If you set the grounding point at the negative pole, the voltage at the plus pole will be 12 V, and the voltage on the other side of R1 will be 8 V.
The voltage difference across R1 is (12 V - 8 V) = 4 V again.

 

[Femme_physics]

Okay, so if the voltage at the + pole is 4V, and it drops 4, and then it drops 8, then the voltage at the - pole is -8, yes?

 

[I like Serena]

Okay, so if the voltage at the + pole is 4V, and it drops 4, and then it drops 8, then the voltage at the - pole is -8, yes?

Yes. :)

 

[Femme_physics]

Must...learn...more! :D

Thanks!

 

[I like Serena]

must...learn...more! :d

thanks!

 

[sauermaische]

You can put 0V anywhere you like in your circuit. The potential that you choose for 0V need not be the same potential that someone else chooses to be 0V in their circuit. When I teach this, I get a pupil to come and stand at the front of the class. I then take a 30m tape measure and declare that I am going to measure the length of their nose. I can choose to put the end of the tape at the top of the nose and measure down; or I can choose to put the end of the tape at the bottom of his nose and measure up; or I can choose to put the end of the tape in the corner of the room and measure the distance to the tip of his nose and to the top of his nose and subtract one measurement from the other. When we use a tape measure we are quite happy that we choose where we put the 0 point. It's the same with a circuit. You get to choose where to put the 0V point. To make things easy we put it at the lowest potential in our circuit (the negative terminal of the battery), but we don't have to. If you put it at the positive terminal of the battery, then all the voltages in your circuit would be negative.

 

[Drakkith]

Hrmmm...I have been working a little bit here and there trying to build a Fusor, a tabletop fusion device, which uses high negative voltage on a grid to ionize gas into a plasma, and then to attract positive ions into the center of the device.

I had assumed that this meant that the applied negative voltage was a buildup of electrons on the grid, while an applied positive voltage was the removal of electrons on the grid, causing a net difference in the number of charges and the resulting force was because of that.

Is my view at all correct?

 

[Dmytry]

The voltage is the electric potential, whereas the number of extra or deficient electrons (times charge of electron) is the charge. You can have a lot of extra electrons at an object, yet it can still be at zero potential, if there's a positively charged object nearby.

The fusor actually has positive potential on the grid (all potentials relatively to the chamber), which means electrons in gas accelerate as they go towards the grid. Inside the grid, the field from the grid itself is close to zero, so the electrons continue their way to the centre. The concentration of electrons increases towards the centre, resulting in increasing negative potential towards the centre (relatively to grid), i.e. the electric field directed towards the centre. The ions are accelerated by this field, colliding in the centre and fusing. The potentials give you the potential energy of the charge brought there. The electric field is the gradient of the potential.

There's a good explanation of how charge density relates to the electric field and potential:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/laplace.html

Ahh, and also.
Inside enclosed conductive chamber, the grid has zero excess or deficit electrons when it is at same potential as the conductive chamber around. The chamber itself may be at arbitrary potential, that does not create any electric field (any gradients in the potential) inside of the chamber (see Faraday cage). So yes, the grid must have deficiency of electrons to be at positive voltage versus chamber.

 

[Drakkith]

Ah ok. I'm glad you added that last bit in. The linked site is almost useless to me since I don't understand any of the math. (Such as whatever that triangle thing means. Delta?)

Edit: Also, I've heard of that version of the fusor, but I was just going to stay simple and use a negative inner grid.

 

[Dmytry]

Upside down triangle is nabla operator... its d/dx, d/dy, d/dz . When you dotproduct or crossproduct or multiply nabla with something, you write it down as if you properly multiplied those components (which is rather crazy). It is very useful when you get hand of it. It describes behaviour of electric field (or anything else really) with differential equations on the properties (equations on partial derivatives, if you wish).
I recommend reading about gradient and divergence... it really helps visualise the fields.

The equations for electric potential are same as idealized small displacements of thin rubber membrane; the charges correspond to forces applied to membrane and field corresponds to slope; the faraday cage corresponds to attaching a rigid horizontal ring (or other shape) to the membrane, inside of which membrane is horizontal.
Other useful analogy is temperature as potential (charge as heat source or sink, electric field as variation of temperature, electric permittivity as thermal conductivity, static case when temperatures stopped changing). The equations are same.
This allows to visualise the fields easily; you can't design anything nontrivial if you have to use computer simulation for every scratchpad attempt, so people learn to sort of imagine how it works and even to be able to sketch the fields with rather good accuracy.

 

[Drakkith]

Ah, ok. Thanks!

 

[Mentoros]

So what happens to the current if we have a -10V supply and a 10 Ohm resistor ?

 

[sauermaische]

What do you mean by "a -10V supply"?

Draw your circuit. Choose which point you want to be 0V. Mark up the potential of the other supply terminal. Apply Ohm's Law.

I presume you have a power supply with a 10 Ohm resistor across it's terminals. Choose one power supply terminal to be at 0V. The other, I presume from your question is at -10V volts . Therefore there is a difference in potential of 10V across the resistor. Conventional current will flow from the more positive supply terminal to the more negative supply terminal (electrons will travel in the opposite direction). If the resistor is metallic and at constant temperature, then Ohm's Law will apply, and the current will have a size of 1 Amp.