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Without replicating the details, an electric current of 1 ampere flowing through a 1mm^2 copper wire appears to correspond to an electron velocity of 0.075 mm/s, i.e. electrons move slower than snails!
As a consequence, the kinetic energy associated with the moving electron charges in the wire appear to be miniscule. By considering the energy delivered per second, we would seem to be referring to power, i.e. watts, which allows a comparison with the definition P=VI. This equation suggests that the power associated with 1 ampere driven by a voltage of 100V would be 100 watts, while the power delivered by the kinetic motion of the charged electrons supporting the 1 ampere current flow appears to be virtually negligible. Therefore, I am trying to better understand the physical process of energy conversion and transfer, as a typical description of electrical current and voltage might take the following form:
- Current describes how many electrons are passing through a wire per second and, as such, high current means lots of electrons are in motion.
- Voltage describes how much energy the electrons carry, high voltage means lots of energy.
The unit of electrical current is the ampere, which resolves to charge per second (I=nq/t), where (n) is the number of free electrons carrying charge. Needless to say there can be a lot of free electrons even in 1 mm^3 of copper, but as indicated, the sum of the kinetic energy associated with even [n] electrons flowing is negligible. On the other hand, the unit of voltage (V) can indeed be resolved into energy per charge, but if the electron does not carry the energy as kinetic energy, how is this energy physically assigned to a charged electron?
Voltage (V) can also be defined in terms of an electric field (E=volts per metre), where this electric field presumably represents potential energy. For example, if I consider the electric field [E=F/q] defines a force (F) on a charge (q), there would be some expectation that this would result in a charged electron acquiring a velocity (v) and therefore kinetic energy, but this form of energy has been shown to be negligible, at least, in terms of the electron current flow. Again, in this context, it seems difficult to explain how the electrons ‘carry’ the energy/power associated with the product of (VI)?
Of course, if the current flows through the wire of an electric motor, it creates a magnetic field that causes physical rotation, which presumably represents the transfer of EM field potential energy into kinetic energy of the motor. However, the interchange between the 2 primary forms of energy, i.e. potential and kinetic, seems less obvious than that of the gravitational potential model. Therefore, I was wondering whether any members have any insight to clarify the issues raised. Thanks
As a consequence, the kinetic energy associated with the moving electron charges in the wire appear to be miniscule. By considering the energy delivered per second, we would seem to be referring to power, i.e. watts, which allows a comparison with the definition P=VI. This equation suggests that the power associated with 1 ampere driven by a voltage of 100V would be 100 watts, while the power delivered by the kinetic motion of the charged electrons supporting the 1 ampere current flow appears to be virtually negligible. Therefore, I am trying to better understand the physical process of energy conversion and transfer, as a typical description of electrical current and voltage might take the following form:
- Current describes how many electrons are passing through a wire per second and, as such, high current means lots of electrons are in motion.
- Voltage describes how much energy the electrons carry, high voltage means lots of energy.
The unit of electrical current is the ampere, which resolves to charge per second (I=nq/t), where (n) is the number of free electrons carrying charge. Needless to say there can be a lot of free electrons even in 1 mm^3 of copper, but as indicated, the sum of the kinetic energy associated with even [n] electrons flowing is negligible. On the other hand, the unit of voltage (V) can indeed be resolved into energy per charge, but if the electron does not carry the energy as kinetic energy, how is this energy physically assigned to a charged electron?
Voltage (V) can also be defined in terms of an electric field (E=volts per metre), where this electric field presumably represents potential energy. For example, if I consider the electric field [E=F/q] defines a force (F) on a charge (q), there would be some expectation that this would result in a charged electron acquiring a velocity (v) and therefore kinetic energy, but this form of energy has been shown to be negligible, at least, in terms of the electron current flow. Again, in this context, it seems difficult to explain how the electrons ‘carry’ the energy/power associated with the product of (VI)?
Of course, if the current flows through the wire of an electric motor, it creates a magnetic field that causes physical rotation, which presumably represents the transfer of EM field potential energy into kinetic energy of the motor. However, the interchange between the 2 primary forms of energy, i.e. potential and kinetic, seems less obvious than that of the gravitational potential model. Therefore, I was wondering whether any members have any insight to clarify the issues raised. Thanks