P=U*I: Why do electrons drop ALL energy e*U in the circuit?

In summary: The power consumed by the load is calculated by P=U\cdot I=U\cdot e\cdot n=\Delta E\cdot n.In this computation, we assume the electron gives all its energy to the load and has kinetic energy zero when it arrives at the plus pole of the power source. But why is that? Why can't the electron maybe only lose half its energy to the load and still have kinetic energy when it enters the battery?The kinetic energy of the electrons is pretty much irrelevant for any device other than a particle accelerator.However, in a bigger picture view of the question, you should not think of energy being stored in individual electrons to be dropped off elsewhere later. The energy is in the
  • #1
greypilgrim
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Hi.

Consider a simple circuit consisting of a voltage source ##U## and a load with resistance ##R##, e.g. a lamp or a motor. The current is given by ##I=U/R##. The number of electrons passing the circuit per second is ##n=I/e##. The power consumed by the load is calculated by
$$P=U\cdot I=U\cdot e\cdot n=\Delta E\cdot n\enspace,$$
where ##\Delta E=U\cdot e## is the (kinetic) energy an electron would gain traveling from the negative to the positive pole of the power source if there was no load.

In this computation, we assume the electron gives all its energy to the load and has kinetic energy zero when it arrives at the plus pole of the power source. But why is that? Why can't the electron maybe only lose half its energy to the load and still have kinetic energy when it enters the battery?
 
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  • #2
The kinetic energy of the electrons is pretty much irrelevant for any device other than a particle accelerator.

However, in a bigger picture view of the question, you should not think of energy being stored in individual electrons to be dropped off elsewhere later. The energy is in the fields.
 
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  • #3
But how does the electron know this? Why does it "think" while passing the lamp "I need to give all my energy to the filament such that I arrive at the battery at rest"?
 
  • #4
The electron doesn't know that. Neither does the electron carry energy to give to a load. The fields carry the energy.
 
  • #5
Consider what happens when you turn on a light switch. The change in the fields moves at a little less than the speed of light. By contrast, the electrons move at about a mm/s. How quickly does the energy get from the source to the light? Is it something that happens nearly at the speed of light or closer to a mm/s?
 
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  • #6
Ok, then I think I need a different derivation of ##P=U\cdot I##. Because the equation I wrote in #1 starting from the right is basically the derivation I learned in school, i.e. looking at the energy difference of an electron (or any other current carrier) between the poles of a voltage source and then counting how many of them pass the circuit per second.

How can ##P=U\cdot I## be derived alternatively without looking at individual charges and make assumptions about how much energy they drop in the circuit?
 
  • #7
greypilgrim said:
How can P=U⋅I be derived alternatively without looking at individual charges and make assumptions about how much energy they drop in the circuit?
In terms of circuit theory it should not be derived, it should simply be defined. Then with KVL and KCL you can show that it leads to energy conservation.

The place where you would derive it would be in electromagnetism, after Poynting's theorem is introduced. I like the treatment here, in chapter 11, especially 11.3

http://web.mit.edu/6.013_book/www/book.html
 
  • #8
greypilgrim said:
Consider a simple circuit consisting of a voltage source U and a load with resistance R, e.g. a lamp or a motor. The current is given by I=U/R.
The current drawn by a motor will be less than V/R unless the motor shaft is prevented from turning. Also, the voltage across the load should be used, not the voltage of the source.
 

Related to P=U*I: Why do electrons drop ALL energy e*U in the circuit?

1. What is the meaning of P=U*I in circuitry?

P=U*I is known as the power formula in circuitry, where P represents the power in watts, U represents the voltage in volts, and I represents the current in amperes.

2. How does P=U*I explain the energy transfer in a circuit?

P=U*I explains the energy transfer in a circuit by showing that the power (P) is equal to the product of the voltage (U) and the current (I). This means that the energy is transferred from the source (voltage) to the load (current) in the circuit.

3. Why do electrons drop all of their energy e*U in a circuit?

Electrons drop all of their energy e*U in a circuit because of the conservation of energy principle. This means that the total energy of the system (in this case, the circuit) must remain constant. As the electrons flow through the circuit, they lose energy due to resistance and this energy is converted into other forms such as heat or light.

4. Can P=U*I be used for both AC and DC circuits?

Yes, P=U*I can be used for both AC (alternating current) and DC (direct current) circuits. However, for AC circuits, the values of P, U, and I are constantly changing due to the alternating direction of the current flow.

5. How does P=U*I relate to the efficiency of a circuit?

P=U*I is directly related to the efficiency of a circuit. The efficiency of a circuit is given by the ratio of the output power to the input power, and P=U*I represents the output power. Therefore, the higher the value of P=U*I, the more efficient the circuit is at transferring energy from the source to the load.

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