# Potential Difference

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#1
Hello Everyone,

I am studying circuits at the moment at GCSE and I quite confused by some areas. I think I have a general understanding of a battery suppling voltage (like a pressure) to the circuit to move the electrons around and that this potential difference is generated because electrons are removed from atoms using chemical energy which is then converted to potential energy whereby the electrons what to recombine with the ions and this is what drives them around the circuit. I then understand that the electrons then travel through components of the circuit i.e. a bulb and this energy is then transferred to the bulb as work is done by the resistance of the bulb however, my question is if we have a 1.5V battery that supplies 1.5J/C and then a lamp that uses 1.5V won't this remove all of the supplied electrical potential as it transferred into thermal energy and light. What would happen to the circuit? Do the electrons just stop flowing? Any help is appreciated!

Thanks!
Last edited by Confusedlady29; 7 months ago
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7 months ago
#2
The electrons don't just stop, if you measured the current before the lamp and the current after the lamp you'd find the same current. The electrons don't build up inside the copper wire, the copper wire is full of electrons to start with and the number of electrons going in has to be marched by the number coming out.

Current is Coulombs per second and Voltage is Joules per coulomb so the bit of the circuit where the work Is being done is the bit with the pd across it.
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#3
(Original post by Joinedup)
The electrons don't just stop, if you measured the current before the lamp and the current after the lamp you'd find the same current. The electrons don't build up inside the copper wire, the copper wire is full of electrons to start with and the number of electrons going in has to be marched by the number coming out.

Current is Coulombs per second and Voltage is Joules per coulomb so the bit of the circuit where the work Is being done is the bit with the pd across it.
Thanks Joined up, Okay I'm still a little confused. As I'm seeing it as the PD input by the battery basically gives energy to the electrons (6.24X10^18) electrons as you say in joules per coulomb. So I'm on the lines of okay let's say we have a 1.5V battery this would give 1 coulomb 1.5 joules by definition. If this was to pass through a lamp of 1.5V then this lamp would effectively use 1.5J/C of the PD so if you measured in parallel to it you would see 1.5V on the voltmeter. This would leave you with no PD left surely as it's all been transferred to the lamp? If this is so then the voltage is what pushes the electrons around surely this 'pushing force' would be reduced after the lamp in the circuit to be zero?
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7 months ago
#4
Hello Everyone,

My question is if we have a 1.5V battery that supplies 1.5J/C and then a lamp that uses 1.5V won't this remove all of the supplied electrical potential as it transferred into thermal energy and light. What would happen to the circuit? Do the electrons just stop flowing? Any help is appreciated!

Thanks!
Your question got me thinking for a while!!

After a fair amount of research online, what I've found is that the idea of an electron 'giving up' energy to a lamp is somewhat crude. At a higher level, the prevailing idea is that charge carriers/electrons aren't necessarily carrying energy per particle. However, I don't understand these convoluted concepts fully myself so I can't give you a better explanation.

As for what you said about 'electrons just stopping to flow' due to a lack of energy. Do note that by placing an electron in between a potential difference, you don't need to provide any energy to get the electron moving from negative to positive, in a similar way you do not need to provide any energy to a ball that falls to the ground. An electron (again very crudely speaking) is "attracted" to the positive terminal and hence this "pulling force" is what causes an electron to move towards the positive terminal, even if it "loses all its energy" to the lamp.
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7 months ago
#5
Thanks Joined up, Okay I'm still a little confused. As I'm seeing it as the PD input by the battery basically gives energy to the electrons (6.24X10^18) electrons as you say in joules per coulomb. So I'm on the lines of okay let's say we have a 1.5V battery this would give 1 coulomb 1.5 joules by definition. If this was to pass through a lamp of 1.5V then this lamp would effectively use 1.5J/C of the PD so if you measured in parallel to it you would see 1.5V on the voltmeter. This would leave you with no PD left surely as it's all been transferred to the lamp? If this is so then the voltage is what pushes the electrons around surely this 'pushing force' would be reduced after the lamp in the circuit to be zero?
The energy isn't the the kinetic energy of the electrons, the electrons aren't starting out fast and slowing down as they go round the circuit.
sticking to the watery analogy, in something like a hydro electric power station the water that's done the work of moving the turbine has to move out of the turbine to make way for the water still coming down the pipe beind it. The rate of water flow into the turbine is the same as the flow out of the turbine. The water isn't building up inside the turbine.
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#6
(Original post by Joinedup)
The energy isn't the the kinetic energy of the electrons, the electrons aren't starting out fast and slowing down as they go round the circuit.
sticking to the watery analogy, in something like a hydro electric power station the water that's done the work of moving the turbine has to move out of the turbine to make way for the water still coming down the pipe beind it. The rate of water flow into the turbine is the same as the flow out of the turbine. The water isn't building up inside the turbine.
Hi Joined up, Thanks for replying. What you have said above makes sense but in that case I am still not getting energy transfer fully then because in my mind some of the GPE energy from the water would have mechanically done work on the turbine and turning into KE and therefore the GPE store of the water would be less so I'd think it would flow at a slower rate. How come this is not the case? As energy is not created nor destroyed but can only be transferred....
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7 months ago
#7
Hi Joined up, Thanks for replying. What you have said above makes sense but in that case I am still not getting energy transfer fully then because in my mind some of the GPE energy from the water would have mechanically done work on the turbine and turning into KE and therefore the GPE store of the water would be less so I'd think it would flow at a slower rate. How come this is not the case? As energy is not created nor destroyed but can only be transferred....
Yes, you are right. The speed of the electron does decrease as it passes through a resistor. Say for example, the resistor (on the atomic level) has densely packed layers of atoms. Imagine on electron trying to cross the resistor, it will have to inevitably collide with the atoms of the resistor. By doing so, it loses some of the electrical energy in the form of heat (hence a resistor would warm up). This is how the concept of 'mean drift velocity' comes into play if you are familiar with it.

However, the electron does not "stop" by losing all of it's energy as you had earlier proposed. This is because the electron is placed in between a potential difference, and so would be "attracted" to the positive terminal, as I'd described in a post above.
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7 months ago
#8
(Original post by qwert7890)
Yes, you are right. The speed of the electron does decrease as it passes through a resistor. Say for example, the resistor (on the atomic level) has densely packed layers of atoms. Imagine on electron trying to cross the resistor, it will have to inevitably collide with the atoms of the resistor. By doing so, it loses some of the electrical energy in the form of heat (hence a resistor would warm up). This is how the concept of 'mean drift velocity' comes into play if you are familiar with it.

However, the electron does not "stop" by losing all of it's energy as you had earlier proposed. This is because the electron is placed in between a potential difference, and so would be "attracted" to the positive terminal, as I'd described in a post above.
I don't agree with this.

If the resistor is made of a thin wire (like the filament in a lamp) and the conductors are made of thick wires then the drift velocity would have to be higher in the lamp filament. because the same number of electrons must be going into the filament per second as are coming out the opposite end but the cross sectional area of the thin wire is lower.

It's more correct to think of the electrons in metal doing a random walk when there's no current flowing and the same random walk with a slight bias in one direction when there's a current flowing - that is what produces the net flow of current.
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7 months ago
#9
(Original post by Joinedup)
I don't agree with this.

If the resistor is made of a thin wire (like the filament in a lamp) and the conductors are made of thick wires then the drift velocity would have to be higher in the lamp filament. because the same number of electrons must be going into the filament per second as are coming out the opposite end but the cross sectional area of the thin wire is lower.
I may be wrong but this is a direct quote from my A Level book:

”It may surprise you to find that, as suggested by the result of Worked example 3, electrons in a copper wire drift at a fraction of a millimetre per second. To understand this result fully, we need to closely examine how electrons behave in a metal. The conduction electrons are free to move around inside the metal. When the wire is connected to a battery or an external power supply, each electron within the metal experiences an electrical force that causes it to move towards the positive end of the battery. The electrons randomly collide with the fixed
but vibrating metal ions. Their journey along the metal
is very haphazard. The actual velocity of an electron between collisions is of the order of magnitude 105 m s−1, but its haphazard journey causes it to have a drift velocity towards the positive end of the battery. Since there are billions of electrons, we use the term mean drift velocity v of the electrons.”
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7 months ago
#10
(Original post by qwert7890)
I may be wrong but this is a direct quote from my A Level book:

”It may surprise you to find that, as suggested by the result of Worked example 3, electrons in a copper wire drift at a fraction of a millimetre per second. To understand this result fully, we need to closely examine how electrons behave in a metal. The conduction electrons are free to move around inside the metal. When the wire is connected to a battery or an external power supply, each electron within the metal experiences an electrical force that causes it to move towards the positive end of the battery. The electrons randomly collide with the fixed
but vibrating metal ions. Their journey along the metal
is very haphazard. The actual velocity of an electron between collisions is of the order of magnitude 105 m s−1, but its haphazard journey causes it to have a drift velocity towards the positive end of the battery. Since there are billions of electrons, we use the term mean drift velocity v of the electrons.”
Yeah, at A level you'll probably be using the formula I=vnAq on different materials and that gives a result that most people find surprising...

The drift velocity in good conductors like copper is really quite low and the drift velocity in poor conductors is much higher - the electrons aren't slowing down to get through resistors.

And obviously you don't need to wait for electrons to move from the battery to the bulb before the bulb starts to glow - the electrons start coming out of the wire at the bulb end (almost) the same instant that other electrons are getting into the wire at the battery end.
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#11
Thanks everyone,

Some of this is a little to complex for what I am currently studying. I think I have now understood it because V= IR so therefore the potential difference component is made from both Current and Resistance parts. So I see this as the resistance part is doing work to warm up the light element and this means that the current is always flowing as that part of the equation cannot be Zero as this would lead to no potential difference therefore the current must always be flowing? Is that correct?
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