# Confusion about voltage in a circuit

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1. Feb 17, 2015

### x86

So basically, I am inferring the following based off of what I've learned.

When we hook a battery up to a circuit, this creates a potential difference. Say we use a 9V battery. Here is a picture:

From what I know, the horizontal is a equipotential surface, and the vertical has different voltages associated with different points. For example, the top point is 9V, middle 4.5V, bottom 0V.

I have some questions:

(1) Now, let's assume we placed a capacitor on the circuit in the position of the 4.5V dot. Then why do we say the voltage across the capacitor is 9V? Shouldn't the voltage across the capacitor be very tiny, depending on the capacitors size? I.e., the voltage at one plate minus the voltage at the other plate? This certainly wouldn't be 9V.

(2) When we have electrical components on the circuit- why is it that we can say they consume voltage and cause a voltage drop? Isn't this impossible? Isn't it defined that the top is 9V, the bottom is 0V. Therefore, shouldn't it be impossible for this voltage to drop, unless we change the batteries voltage?

(3) If the red line is an equipotential surface, exactly what encourages the current to flow? Is it that the electrons will gather up along this equipotential surface, and eventually end up repelling each other? Thus causing the current to flow?

2. Feb 17, 2015

### davenn

That's incorrect .... Why would you think that ?

You have not indicated the presence of a resistor, so assuming you are indicating a normal piece of copper wire, there will be a short circuit across the battery and the voltage will be zero or something close to it

incorrect ... see my comments above

No, we don't say, "they don't consume voltage".

again, refer to my above comments

no ... again refer to my first comments

Dave

3. Feb 17, 2015

### jim hardy

Edit: oops - i see Dave already answered while i was typing

sorry, guys..

X86:
Have you really thought about the definitions of potential and voltage? You seem to be mixing the concepts.
Nobody is going out to infinity to make that measurement for us.
So we always use the difference of potential between two local points.
That's the definition of voltage, potential difference between two points..

So whenever you state a voltage you must make it clear between what two points.

I wouldn't say that at all.
A capacitor has two ends not one. Where do they go? You only mentioned one place. It takes two places to have a potential difference.

old jim

Last edited: Feb 17, 2015
4. Feb 17, 2015

### davenn

The ONLY way you would get voltage reading like you have indicated would be if you had 2 resistors of identical value in series between the positive and negative terminals of the battery
like this .....

so ...
--- if you put the negative lead of your voltmeter at point C and the positive lead on point A, you would measure 10V
--- if you put the negative lead of your voltmeter at point C and the positive lead on point B, you would measure 5V
--- if you put the negative lead of your voltmeter at point B and the positive lead on point A, you would measure 5V

5V is dropped across each resistor
This is called a voltage divider. if the resistances are identical, the centre voltage at point B will always be 1/2 the full supply voltage
if the resistor values are different, then the voltage at point B will be some value relating to the voltage drops across each resistor

Dave

Last edited: Feb 17, 2015
5. Feb 17, 2015

### davenn

Now there is one proviso (caveat) to this
I wont mention it yet and confuse you, I want to make sure you have got the basics that Jim and I have stated above
once those are sorted I will comment on the other "option"

Dave

6. Feb 18, 2015

### x86

I think after reading the above posts, I now understand where my confusion was coming from. I know that voltage is potential difference (we must measure the potential in regards to two points)

But I'm a little confused about one concept regarding the below circuit diagram:

a) What is the potential difference betwen A and B? Isn't it zero, because it's an equipotential surface?
b) What is the potential difference between B and C?
c) What is the potential difference between B and D?

I'm thinking b and c have the same answer, but this is kind of confusing to me, because they have different lengths.

This also kind of relates to my capacitor confusion. If we put a capacitor at C then the potential difference between its plates will be 9V. But why? This part confuses me, because it feels like B--->C should have a different potential difference than B--->D (related to the wire length)

7. Feb 18, 2015

### davenn

hi x86

Am not sure why you are still confused ? did you study that circuit I posted with the 2 resistors ?

did you not understand the text I posted at the top of my first reply ?.....

in your latest diagram, there is no difference between any of those points ( when dealing with a length of copper wire a few metres or so long )

IM thinking your confusion stems from the fact that part of the wire is level and part if vertical .... is that correct ?

if so you are making incorrect assumptions

Dave

Last edited: Feb 18, 2015
8. Feb 18, 2015

### x86

Yes. I did read your above post. I am just confused about the voltage between different points in the above diagram (copper wire only).

I still don't understand what you mean by "there will be a short circuit across the battery and the voltage will be zero or something close to it."

My confusion is stemming from being unsure what the potential difference in a battery hooked up to a copper wire is, at different points.

For instance, lets say that A and B have different potentials (somewhere in space). Then their potential difference is B-A

But how does this relate to the circuit? Now the points A,B are no longer in space, but on the circuit. I know the battery provided potential difference. But I'm confused about the potential difference on different parts of the circuit.

9. Feb 18, 2015

### davenn

OK

consider a short to say a few metres length of copper wire ... it's going to have close to zero Ohms resistance ( copper is a GOOD conductor)

if you put that bit of wire across your battery, its going to short circuit the battery, the wire is going to get VERY hot and the battery is going to go flat quickly
as there will be a large current flowing Even putting your voltmeter directly across the battery terminals you will read zero ( of something very close to) volts

BECAUSE that length of wire is very low resistance, ALL the battery voltage will be dropped across that bit of wire

Now here's part of that proviso I hinted at earlier .....

That bit of copper wire ISNT a perfect conductor it has a VERY SMALL resistance over its length. therefore if you have a good voltmeter that can read small voltages, you will likely read a difference in voltage between 2 points, but it will only be a few micro to milliVolts depending on the length of the wire and how apart your measurement points are

But for practical purposes, for a length of wire under a few metres, you don't need to take that into consideration and it can be ignored, as other components in the will produce much larger voltage drops

Last edited: Feb 18, 2015
10. Feb 18, 2015

### davenn

have a look at this pic I drew .....

The voltage measured between points A and B in any of those 3 circuits will ALL be 10V

11. Feb 18, 2015

### jim hardy

A question well stated is half answered.

You are sending us mixed messages.
Your horizontal wire AB you say is an equipotential surface.
Then you say your vertical wire is not an equipotential surface.
What is the difference between those two wires? Electric potential is unaffected by gravity.

If there is no potential difference between A and B there'll be no current flow from A to B.
If AB BC and CD are all made from the same kind of wire, and are as near the same length as they appear in your drawing,
then why do they (edit have) behave so differently when connected in series?

In circuit analysis we usually assume our conductors to have zero resistance hence zero voltage drop.
But we remain aware that's an exaggeration, in actuality there is small voltage required to push current along a wire.
I've measured the millivolts along printed circuit tracks to see where current was going.
In power distribution we aim for less than 3% voltage drop in our feeder.

Now - a 9 volt transistor radio (006P) can make only about an amp or two
i'd expect an amp or two to flow
the battery to get very hot
and potential difference of only millivolts per inch along your wire.

If you apply enough current to drop volts per inch along the wire it will quickly glow red hot and melt. That's how a soldering gun works.

Dave has given great direction.
You need only to slow down and realize that you already know these answers.
We live in such a rushed and instant-response world that it is difficult to make ourselves think in the small incremental steps requisite for figuring things out.
Much of learning is discovering what we already know.

Once again you have failed to define to what two points your capacitor's two plates are connected.
Which plate is "put at C"?
Where's the other plate connected ?
How can you conceive of a potential difference between only one point?
It is unclear what you are thinking.

"Slow down, you move too fast" simon&garfunkel

Last edited: Feb 18, 2015
12. Feb 18, 2015

### x86

Thank you guys for being so patient with me, and I am sorry for sending mixed signals. After pondering for a while, I've realized that I've been unnecessarily overcomplicating the problem. Let's assume the circuit is a superconductor and the wire has a resistance of 0 V.

You are correct, I am confused about the fact that some of the wire is level and some of the wire is vertical. I was told that in circuits, the level wire will always be an equipotential surface, and that the vertical wire will always have a voltage equal to that of the battery. I do not understand why this is the case. The only thing that I know is that it will take 9J of work to move a charge of 1 C from the (+) terminal to the (-) terminal. Everything I said in my first post were things that I think, not necessarily things that are true.

13. Feb 18, 2015

### davenn

its OK

As Jim said ... and one of his fav saying on the forums .... "A question well stated is half answered"

another words, if you sit and think about how to ask a good question ( well constructed), chances are you already have a good part of your answer

If you don't mind me asking, what is your age and what school level are you at ?
It can be difficult on a forum like this sometimes, because we cant see the person or know their background education
its sometimes difficult to know how easy/complex to make the answers

cheers
Dave

14. Feb 18, 2015

### davenn

That is totally incorrect and no wonder you were getting confused with our responses.

think about the wiring in just about any electronics .... inside a TV, a computer, a car, a plane

the wiring curls all over the place as it makes its way from power sources to control panels/displays to the sensors etc being controlled.
The potentials are not affected by that at all. What affects the voltage drop on a wire is its length.
As the wire gets longer, its resistance increases and this produces a voltage drop

D

15. Feb 18, 2015

### x86

Yes, I apologize if my questions are not well stated. I really need to work on my communication skills. I am a first year computer engineering student taking an introduction to electricity course. Basically, I learned about the basics such as coulombs law, electric fields, gausses law, amperes law, capacitors, electric flux, magnetic flux, electric potential,capacitance, dialectrics, current, resistance, current density, magnetic fields, induced magnetic fields, faradays law, lenz law. And a few other basic stuff. It is probably a very basic physics course, using calculus. Now we are on circuit analysis.

16. Feb 18, 2015

### davenn

cool ... lots of good learning ahead of you

never be afraid of asking for help, all of us on here do our best to help, but what we really like doing is teaching people to help themselves
ie. guide you so you can work out the answers for yourself ( but sometimes just giving a straight answer to clear confusion works well)

17. Feb 18, 2015

### x86

Ah okay. Thank both of you for your help, and sorry about the confusion. It is actually pretty simple when I think about it. It takes 9 J to pump a 1 C charge from the (+) terminal to the (-) terminal in the above diagram, and if the wire has resistance, then we lose some of this energy to heat, and of course how much we lose depends on the length of the wire and the resistance of the wire.

18. Feb 18, 2015

### davenn

from the (-) terminal to the (+) terminal

electrons are the charge carriers and they move from negative to positive

Have you been told about the difference between conventional current flow + to - and electron flow - to + and why there is a difference ?

Dave

19. Feb 18, 2015

### x86

Yes. Current is how a positive charge would flow in the circuit. It moves from high potential to low potential, whereas electrons move from low to high potential. Of course, the only thing flowing is electrons, so current seems like something more imaginary. In my example above, I was referring to an imaginary +1C charge.

20. Feb 18, 2015

### jim hardy

That is good drafting procedure to draw them that way and nothing more.
In vacuum tube days we drew + at top of page and signal flow left to right, top to bottom just like we read..

Today's computer drawings are nightmares- signals go every which way.

Lots of people do.
I quote this guy below often, so do not be offended if i seem repetitious.
Forty five years ago i bought a book at a library yard sale.
The first essay in it was Lavoisier's "Introduction to his treatise on Chemistry".
I started it and found, though the language seems archaic (translated from 1700's French)
it is a great primer on analytical thinking and clear communication.
Over the years i came to appreciate just how rare those skills are. But they can be learned, in fact my company gave courses in "analytical troubleshooting".

Pay special attention to his quote near the end starting with "Instead of applying observation to things we wish to know..."
you'll see it a lot.
Here it is:
https://web.lemoyne.edu/giunta/lavpref.html

Here's the last line:
and that's why engineers need to take some English credit hours. (English speaking engineers, that is)

good luck, happy to see you working your way out of confusion.
thanks for the kind words.

old jim