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Energy And Electric Field In Circuits

  • Thread starter didas
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  • #1
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Hi there,

Sorry if this is in the wrong section.

I'm currently doing some work on models for electric circuits, but my own are not particularly complete! If anyone could shed some light on the following, I would be very grateful:

How do the electrons transfer the energy in a circuit? I'm fully up to speed with e.m.f. and p.d. and I can derive drift velocity, but the mechanism for the transfer of energy eludes me. I tried to work out the total kinetic energy of the electrons in a wire based on their drift velocity, but this was massively lower than the energy transferred to the charge in the same wire. I'm particularly intrigued as to why P=I^2R. If we double the current we quadruple the power, which would seem possibly to be related to kinetic energy as K.E.=1/2mv^2. I'm sure this is down to something quantum, so if it is, please be gentle with me!

Secondly, how does the electric field vary within a series circuit? I'm looking for the reason as to why the p.d. is larger across resistors than it is across wires. I understand we can predict this using V=IR or through the potential divider formula, but I'm interested in the mechanics of it. What is it about a resistor compared to a conductor that creates a larger p.d. across it (and I guess a stronger electric field through it).

Any help would be very welcome, thanks.

Graham
 

Answers and Replies

  • #2
Andrew Mason
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How do the electrons transfer the energy in a circuit? I'm fully up to speed with e.m.f. and p.d. and I can derive drift velocity, but the mechanism for the transfer of energy eludes me.
The electrons acquire kinetic energy from the electric field. Now, how they transfer that energy depends on the device to which the electric field is applied. If the electric field is applied to a light bulb, the electrons in the light bulb filament gain kinetic energy and start crashing into the tungsten atoms in the filament and cause them to vibrate, raising the temperature of the filament and causing it to glow. If the device is an electric motor, the electron kinetic energy is converted into magnetic field energy that causes the armature to rotate.

I tried to work out the total kinetic energy of the electrons in a wire based on their drift velocity, but this was massively lower than the energy transferred to the charge in the same wire. I'm particularly intrigued as to why P=I^2R. If we double the current we quadruple the power, which would seem possibly to be related to kinetic energy as K.E.=1/2mv^2. I'm sure this is down to something quantum, so if it is, please be gentle with me!
The fact that electrons have a small drift velocity does not mean they do not acquire high kinetic energy. They do. It is just that they keep crashing into atoms.

P=I^2R because V = IR. V is the potential difference per unit charge. So the energy of an electron to which such a potential difference is applied is qV. So the power, the energy per unit time, is d/dt(qV) = IV = I(IR).

Secondly, how does the electric field vary within a series circuit? I'm looking for the reason as to why the p.d. is larger across resistors than it is across wires. I understand we can predict this using V=IR or through the potential divider formula, but I'm interested in the mechanics of it. What is it about a resistor compared to a conductor that creates a larger p.d. across it (and I guess a stronger electric field through it).
A copper wire has a low resistivity. It does not take much energy to move electrons through a copper wire. A carbon resistor has a high resistivity. It takes more energy to make electrons move in a carbon resistor. So, in moving a given current through a circuit of copper wire and a carbon resistor, more electrical potential will be needed for the resistor than for the copper wire.

AM
 
  • #3
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Thanks for your reply Andrew, very kind. However, I'm not sure we've really got to the bottom of my question. As I mentioned, I worked out the KE of all of the electrons in a piece of wire carrying a certain current using their drift velocity and the charge carrier density for copper. For the same p.d. that would produce the current I used W=QV to calculate the energy transferred to the charge and it was many orders of magnitude out. I don't have my scribblings to hand right now, but I'll post them in the morning (it's getting late here now).

Also, I'm really after the mechanism that causes there to be a stronger electric field in a resistor. You say that more electric potential will be needed and yes, I understand that, but using "needed" sounds like the circuit has a mind of its own! What causes a stronger electric field to occur in a resistor, rather than a conductor?
 
  • #4
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Energy is not transferred kinetically by the electrons. Otherwise a conductor using protons would transfer more right? But it wouldn't.

The energy is transferred by the charges moving with or against the fields, both electric and magnetic.
 
  • #5
Andrew Mason
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Thanks for your reply Andrew, very kind. However, I'm not sure we've really got to the bottom of my question. As I mentioned, I worked out the KE of all of the electrons in a piece of wire carrying a certain current using their drift velocity and the charge carrier density for copper. For the same p.d. that would produce the current I used W=QV to calculate the energy transferred to the charge and it was many orders of magnitude out. I don't have my scribblings to hand right now, but I'll post them in the morning (it's getting late here now).
You cannot simply take all the available charge carriers in a piece of copper and multiply that by the potential difference that is applied. Current does not work that way. At the microscopic level, there are all sorts of non-classical things happening. These are all bound up in the concept of resistivity.

You divide the potential difference by the resistance to get the current - which is, in the classical concept, the number of unit charges that flow pass any given point in the circuit per unit time.

This concept of current was developed long before we understood quantum mechanics. At the atomic and electron level, there is not a fixed number of charges passing a given point per unit time at a particular speed. The actual velocity of the charges is very difficult to measure because they keep bouncing around randomly inside the conductor and, on average, move rather slowly. They are also not distinguishable. So don't think of current as charges moving with a certain emf-driven velocity through a conductor.

Also, I'm really after the mechanism that causes there to be a stronger electric field in a resistor. You say that more electric potential will be needed and yes, I understand that, but using "needed" sounds like the circuit has a mind of its own! What causes a stronger electric field to occur in a resistor, rather than a conductor?
The current self-adjusts so that the voltage drop across the resistor is proportional to the size of that resistance relative to the rest of the circuit. Think of a pipe of water with different constrictions in series. The one with narrowest constriction will have the greatest pressure drop. This is because the narrower the constriction the greater the reduction in total flow. Since the rate of energy loss is the pressure drop x flow (P x dV/dt), the greatest energy loss is across the narrowest constriction.

Similarly, with current the greatest resistance is the greatest limiter of current.

Here is another way of looking at it. Consider a 1 metre long cylindrical resistor with uniform resistivity [itex]\rho[/itex]. The resistance is a linear function of distance along the resistor. The current is the same at all points in the resistor. So the potential drop per unit distance along the resistor will be uniform all along the resistor. So, if you mark off the resistor into 10 cm, 30 cm and 60 cm lengths, how will the potential drops across each compare. Now cut the resistor into separate pieces and connect them together in series with wires with negligble resistivity. Will that change anything?

AM
 
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  • #6
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Thanks again Andrew.

So, you are saying that the energy transferred to the electrons in a circuit is entirely as kinetic energy, but that it is not just as the extra kinetic energy caused by the drift velocity, if I understand correctly.

Regarding the electric field/p.d. question. It seems like we have a chicken and egg situation here. The series current and differences in resistance lead to the potential drops, but surely the current is caused by the p.d. and not the other way around. From what I understand, the electric field in a circuit is propagated far more quickly than the electrons travel, so what is it about the material from which the resistor is constructed that concentrates the field?
 
  • #7
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Energy is not transferred kinetically by the electrons. Otherwise a conductor using protons would transfer more right? But it wouldn't.

The energy is transferred by the charges moving with or against the fields, both electric and magnetic.
I think a conductor using protons would transfer more energy at the same current. Conversely, if we accelerated electrons and protons in a vacuum across an equal (but opposite) p.d., they would both aquire the same kinetic energy, but the electrons would reach a higher speed and create a larger current.
 
  • #8
Andrew Mason
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Thanks again Andrew.

So, you are saying that the energy transferred to the electrons in a circuit is entirely as kinetic energy, but that it is not just as the extra kinetic energy caused by the drift velocity, if I understand correctly.

Regarding the electric field/p.d. question. It seems like we have a chicken and egg situation here. The series current and differences in resistance lead to the potential drops, but surely the current is caused by the p.d. and not the other way around. From what I understand, the electric field in a circuit is propagated far more quickly than the electrons travel, so what is it about the material from which the resistor is constructed that concentrates the field?
It is not a matter of the material from which the resistor is made. It is a simply that of the proportion of the applied voltage that is required to maintain a current through that resistance has to be in proportion to the magnitude of the resistor relative to the resistance of the entire circuit. If the total resistance decreases, the current increases (applied V remaining the same). If the total resistance increases, current decreases. In all cases, the sum of the voltage drops across all resistors equals the total applied voltage (assuming only series resistance in the circuit). That has to be the case. If the voltage drops across each resistor were not proportional to the resistance, you would have to have different currents in each, which cannot occur in a series circuit.

AM
 
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