Kinetic Energy and Electron Flow Speed

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I'm learning about electricity and I've stumbled upon a question which I cannot seem to find the answer to. I realize that I am probably missing some key piece of information or making some form of understanding error but just can't seem to grasp what it is.

My understanding is that if you have a current, I then 2I will have an average electron drift velocity that is twice that of I. Now, to produce 2I from I with uniform resistance requires a voltage that is also twice the original, V and 2V respectively.

Kinetic Energy is 1/2mv^2. An two-fold increase in speed will produce a four-fold increase in energy. How is it then that double the potential energy will produce four-times the kinetic energy?
 

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Vanadium 50
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. An two-fold increase in speed will produce a four-fold increase in energy.

So will simultaneously doubling the voltage and the current.
 
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But how do the electrons flow at twice the speed to begin with? The new voltage imparts a double in force. If that force is applied over the same distance, it's only enough to impart a velocity increase by a factor of sqrt(2). If the electrons are only sqrt(2) times faster than before, how can the current be twice as high?
 
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I'm learning about electricity and I've stumbled upon a question which I cannot seem to find the answer to. I realize that I am probably missing some key piece of information or making some form of understanding error but just can't seem to grasp what it is.

My understanding is that if you have a current, I then 2I will have an average electron drift velocity that is twice that of I. Now, to produce 2I from I with uniform resistance requires a voltage that is also twice the original, V and 2V respectively.

Kinetic Energy is 1/2mv^2. An two-fold increase in speed will produce a four-fold increase in energy. How is it then that double the potential energy will produce four-times the kinetic energy?

This kinetic energy is dissipated in your resistance (wire). The power released is I*V=4*I0V0
 
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This kinetic energy is dissipated in your resistance (wire). The power released is I*V=4*I0V0

Sorry, I'm still not understanding this so let me ask what is confusing me. I suspect one of the following statements will be wrong and that's what is wrong with my understanding.

1. A double voltage is a double increase in the electric field driving the current.
2. Doubling this electric field causes the electrons in the wire to feel twice the force and therefore, have twice the amount of energy.
3. The total energy change in each electron is by a magnitude of 2. A double in energy is equal to a speed that is sqrt(2) times faster than before. But for the current to be double as shown by ohm's law, the electron drift speed has to be twice as much, not sqrt(2) times as much since all other factors are unchanged.

It doesn't seem to add up. The energy each electron gained from the double voltage and the speed gained by each electron from the double voltage are not equal.
 

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