Register to reply 
Joules per Coulomb and the Volt 
Share this thread: 
#1
Mar212, 09:32 AM

P: 66

Hey,
A coulomb is the amount of charge that passes a point through a wire carrying one ampere for one second. Voltage is a measure of electrical potential energy in units of volts or joules per coulomb (energy/charge). Then 1 volt means 1 joule per coulomb; 2 volts mean 2 joules per coulomb, and 5 volts mean 5 joules per coulomb. But what is meant by 2 or 5 volts? If one coulomb contains a set amount of electrons, how can one coulomb have more energy or potential to do work than another coulomb? What is different? 


#2
Mar212, 10:04 AM

P: 628




#3
Mar212, 10:21 AM

P: 66

I understand the potato analogy. The higher the height, the greater the momentum. But why does a potential of 2 volts do more work than 1 volt? Is it the amount of the excess/deficiency of electrons that determines how much energy is expelled?



#4
Mar212, 10:34 AM

P: 628

Joules per Coulomb and the Volt
In effect; to determine the voltage, you perform a 'test' on some of the electrons to see how much energy they release when they undergo a transition to a known [or given/datum] potential. A bit like if you don't know the height you are at, so you pull a potato out of the bag, and time how long it takes to fall from your height to wherever you're dropping the potato to. 


#5
Mar212, 10:58 AM

P: 66

So then when we consider current and resistance, 2 V means that in a wire with 1 ohm of resistance, at a certain point in the wire, there will be 2 coulombs of charge passing that point every second.



#6
Mar212, 11:03 AM

P: 628

Yes, that's right. Was there a question? Each coulomb that passes through a resistor with a 2V potential drop across it will release 2J into that resistor (because 2 V = 2J/C). Two Coulombs will mean 4J of energy is released, through a 2V potential drop.
In fact, it is, again, somewhat the other way around  the resistor, R, is such that it will cause J Joules to be released when C Coulombs pass through it, according to J=C^2 x R 


#7
Mar212, 11:17 AM

P: 66

I see, I think I understand now. So the volt is a standard? 1 V means the energy or work that that the charges have the potential to do through a wire of 1 ohm of resistance, in other words, the amount of energy used up through the resistance.
So when we buy a 9 V battery from the store, this means that the battery is somewhat "guaranteed" to do 9 joules of work through a resistance of 1 ohm and at a rate of 9 coulombs per second...? 


#8
Mar212, 11:23 AM

P: 628

Almost. There is a ^2 term in there. I think I'm overstretching the analogy to explain it, but think of the potato and that the wind resistance is a ^2 term of the motion of the potato falling through the air.
The 9V battery denotes that One Coulomb of electrons passing from one terminal to the other can nominally perform 9J of work. Across a 1 ohm resistor there would be 9 Coulombs of charge per second, so that'd be 81 Joules per second. 


#9
Mar212, 11:36 AM

P: 66

That's where I get stuck. I get the calculation of power. With 9V, 9 coulombs/second with each coulomb being able to do 9 joules of work computes to 81 joules/sec (as you said). I don't get what being done to get the electrons to do more work per coulomb, to get more joules out of each coulomb.



#10
Mar212, 12:47 PM

P: 628

The issue is, in effect, the 'resistor' value that makes this look more complicated than you are seeing it to be. The resistance of an electrical element to the flow of electrons is analogous to the resistance of air to an object passing through it  resistance through air is a v^2 term. I'm trying to draw out this analogy (which I don't think is actually a particularly good one! ) only because I can see you are wanting to conceptualise why the power delivered into a resistance is a function of volts^2. I'm not sure I have a better picture to paint for you than that. Maybe someone else will chime in with a better way to provide a conceptualisation of why this is, without simply pointing to the maths of the thing, but maybe you are better just to stick with the maths, and the reality of those equations: R = V/I Energy = V.C Power = Energy/s = V.C/s = VI so Power = V^2/R = R.I^2 


#11
Mar212, 01:19 PM

PF Gold
P: 7,120

So if you want to relate the voltage on a battery to the charged particles falling inward, the voltage basically tells you, in analogy, how far the charged particle can be pulled inwards and the further you pull it in, the more work done per Coulomb of charge. A battery's voltage basically tells you how capable it is of pulling a current through a circuit that has some resistance. 


#12
Mar212, 01:34 PM

P: 66

Maybe all of my confusion will be eliminated if someone answered this. Quantitatively, does the positive terminal of a battery have the same tendency or force to receive electrons as the negative terminal who is trying to give them away? In other words, does the negative terminal have as many excess electrons as the positive has a deficiency?



#13
Mar212, 01:47 PM

P: 628

'Net neutrality' (or otherwise) does not define 'voltage'. You are best to think of 'voltage' as explicitly the relative potentials between two conductors, i.e. as purely a conceptual quantity of charge, for the purposes of your question. Back to the height analogy again  what is important when you fall off a wall is the height of the wall, not your height above or below sea level. 


#14
Mar212, 05:37 PM

P: 66

While thinking about it, I may have answered my own question.
The voltage between two objects is their measured difference in electric potential. So for example the 5V battery means that the difference in the electric potentials is 5V. The potentials themselves can be anything so long as their difference is 5V. The potential of the cathode could be 8V and the anode could be 3V but the difference is still 5V, so, hence the 5V battery. The electric potential (in volts) is the ratio of the electric potential energy (in joules) and the charge (in coulombs). Am I on the right track? 


#15
Mar212, 05:45 PM

PF Gold
P: 7,120




#16
Mar212, 05:59 PM

P: 66

I think I'm beginning to see the picture. So, everything has a force due to electric charges just as everything with mass attracts everything else with mass due to gravity, and you can't really diminish the electric potential to zero. Is this correct as well?



#17
Mar212, 06:32 PM

Sci Advisor
PF Gold
P: 3,657

Think of volts as pressure on the electrons,,, potential energy not kinetic.
I would be careful with the potato analogy because it leads one's mind naturally to imagine their velocity when they hit the floor. Electrons in a wire have quite low velocity, but can do work by pushing against resistance to make heat, or against a magnetic field to make force as in a loudspeaker or electric motor. It's natural to think of parallels with Bernoulli , but be careful. They're pretty near massless compared to real fluids. old jim 


Register to reply 
Related Discussions  
120 volt to 440 volt 3 phase  Electrical Engineering  14  
12 volt switch from 5 volt supply  Electrical Engineering  7  
12 DC Volt Boost to 15 DC Volt  1 amp  Electrical Engineering  3  
How do you convert 3 Volt Control signal to 5 Volt control signal  Electrical Engineering  5  
Proton volt= electron volt  Classical Physics  3 