Power in electric circuit: apparent contradiction

In summary: Electrical current is the number of charges moving past a point in the circuit per unit of time. This can either be few charges moving a high velocity, or many charges moving at a low velocity. You cannot get the power of...
  • #1
Ott Rovgeisha
76
5
Hi, dear forum members!

It is well known, that if one makes the current carrying wire more thicker, for example 3 times, that means the current goes up also 3 times.

Microscopically that seems plausible, because, when we consider that a cross section is made of atoms, then it is apparent, that making the cross sectional area larger 3 times, means that one makes the number of atoms building up the cross-section also 3 times bigger.

If we consider that every atom has one loose electron, then it is very plausible that the number of electrons accelerating through the cross section per unit time is also 3 times larger. That is equal to saying that the current is 3 times larger.

It also well known that the power that the piece of wire exhibits, is P= Iˇ2 * R.

Mathematically, it all makes sense, until one starts to think about it conceptually.

According to one model, electrons accelerate due to the electric field and collide with the atomic cores.

Well, suppose that the field is uniform and that every electron accelerates the same way (for simplicity).

That means that each electron collides, having a certain amount of energy (mvˇ2)/2.
3 times larger cross-section means 3 times more atoms per cross- section, which means 3 times more electrons colliding.

Since each electron is giving away a certain amount of energy (mvˇ2)/2, that means that the 3 times more electrons should give away 3 times more energy during the same time unit.
Multiplying it by the length of the wire, we can get the entire amount of energy that the wire exhibits during this time unit. It is STILL 3 times larger, NOT 9 times as the equation predicts.

What is wrong with the reasoning?
I have a few ideas myself, but I would very much like to see your opinions on this matter.

Kind regards!
 
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  • #2
Ott Rovgeisha said:
It also well known that the power that the piece of wire exhibits, is P= Iˇ2 * R.
It is also well known that I = V/R.
Decreasing the resistance by one third also increase the current by a factor of 3.
 
  • #3
256bits said:
It is also well known that I = V/R.
Decreasing the resistance by one third also increase the current by a factor of 3.

Yes it is, but this is not what I am asking! Did you read through the reasoning?
Making current larger in THIS case means allowing more charge trough, but does not mean that every single charge carries more energy so that it would cause the SQUARE relationship. There is an apparent problem!
 
  • #4
It sure is what you are asking.
With the same EMF, the new current in the thicker wire with 1/3 the resistance will be now 3 times what is was before.
From P = I^2 R, you now have P = (3I) ^ R/3 = 3IR, in the thicker wire, or 3 times the power of that in the thinner wire.
 
  • #5
Ott Rovgeisha said:
Yes it is, but this is not what I am asking! Did you read through the reasoning?
Making current larger in THIS case means allowing more charge trough, but does not mean that every single charge carries more energy so that it would cause the SQUARE relationship. There is an apparent problem!

But if you make the crosssection of the wire 3 times bigger, R will be a factor of 3 smaller so P only goes up by a factor of 3, so everything comes out fine.
 
  • #6
256bits said:
It sure is what you are asking.
With the same EMF, the new current in the thicker wire with 1/3 the resistance will be now 3 times what is was before.
From P = I^2 R, you now have P = (3I) ^ R/3 = 3IR, in the thicker wire, or 3 times the power of that in the thinner wire.

You should SQUARE the (3I) term, shouldn't you?

It would give you 3Iˇ2 * R not 3IR. Why aren't you squaring the I ?
 
  • #7
willem2 said:
But if you make the crosssection of the wire 3 times bigger, R will be a factor of 3 smaller so P only goes up by a factor of 3, so everything comes out fine.[/QUOTE

Ok, so it turns out that the wording in some textbooks and educators who say that power is proportional to the square of the current, are simply wrong!

They conveniently forget to mention that, it is ONLY true, if they would somehow keep the resistance at a constant value.

Thank you!
 
  • #8
Ott Rovgeisha said:
That means that each electron collides, having a certain amount of energy (mvˇ2)/2.
3 times larger cross-section means 3 times more atoms per cross- section, which means 3 times more electrons colliding.

Electrical current is the number of charges moving past a point in the circuit per unit of time. This can either be few charges moving a high velocity, or many charges moving at a low velocity. You cannot get the power of the circuit by looking at it as electrons with kinetic energy.
 
  • #9
Ott Rovgeisha said:
You should SQUARE the (3I) term, shouldn't you?

It would give you 3Iˇ2 * R not 3IR. Why aren't you squaring the I ?
Look again closely at the calculation. The new resistance for the thicker wire is R/3.
 
  • #10
Drakkith said:
Electrical current is the number of charges moving past a point in the circuit per unit of time. This can either be few charges moving a high velocity, or many charges moving at a low velocity. You cannot get the power of the circuit by looking at it as electrons with kinetic energy.

I am fully aware of that, but if we ONLY make the wire thicker, nothing else, then sure as hell the kinetic energy of the electrons before colliding is roughly the same, because we have not changed the strength of the electric field or the the voltage. You most certainly can calculate the kinetic energy transfer from the electrons to atoms!

The power of the circuit comes DIRECTLY from the electrons that are accelerating due to the electric field IN the wire!
 
  • #11
256bits said:
Look again closely at the calculation. The new resistance for the thicker wire is R/3.
Yes the resistance is indeed 3 times lower, but the current should be squared as well!

Thanks for your reply! The main problem seems to be the misconception some textbooks and teachers have, that power is proportional to the square of the current; it would be, if resistance was kept constant somehow.
 
  • #12
Ott Rovgeisha said:
Yes the resistance is indeed 3 times lower, but the current should be squared as well!

Thanks for your reply! The main problem seems to be the misconception some textbooks and teachers have, that power is proportional to the square of the current; it would be, if resistance was kept constant somehow.

Well, one could easily change the voltage of the power supply to give different currents through the same resistance.

You do have a good point about textbooks and teachers.Textbooks cannot anticipate every question a student can have about a subject, and some do explain in a fashion that is attuned for some students and not for others. Teachers will suffer the same ratings, with some students finding a teacher's method easily understandable, for other students not so. And vice-versa.

Cherios!
 
  • #13
Ott Rovgeisha said:
I am fully aware of that, but if we ONLY make the wire thicker, nothing else, then sure as hell the kinetic energy of the electrons before colliding is roughly the same, because we have not changed the strength of the electric field or the the voltage. You most certainly can calculate the kinetic energy transfer from the electrons to atoms!

I think I misunderstood your original question. You were thinking that in the equation P=I^2R, increasing the cross section of the conductor increases the current, which is true. But you didn't take into account that R was reduced as well, so the equation doesn't predict 9x more power, but only 3x more power.
 
  • #14
Drakkith said:
I think I misunderstood your original question. You were thinking that in the equation P=I^2R, increasing the cross section of the conductor increases the current, which is true. But you didn't take into account that R was reduced as well, so the equation doesn't predict 9x more power, but only 3x more power.

Yes, and like 256bits said, one can easily change the voltage, and keep the resistance constant. If that is the case, then the power difference really would be 9 times higher (if there is no change in resistance due to increased temperature...).

So it turns out again, how important it is to be precise in conceptions about different equations; it is a very dangerous thing to jump into conclusions.
I guess that is the beauty of physics: it gives advice about your everyday life: do not jump into conclusions and be aware of details...

Thank you all, for your replies!
 

1. What is the difference between real and apparent power in an electric circuit?

Real power is the actual power that is consumed by a circuit and is measured in watts. It is the power that is converted into useful work, such as lighting a bulb or powering a motor. Apparent power, on the other hand, is the combination of real power and reactive power. Reactive power is the power that is used to establish and maintain electric and magnetic fields in a circuit, but does not do any useful work. It is measured in volt-amperes (VA).

2. Why does apparent power sometimes appear to be greater than real power in an electric circuit?

This can occur when there is a high amount of reactive power in the circuit. Reactive power can cause the voltage and current to be out of phase, which results in a higher apparent power. This can happen in circuits with inductive or capacitive components, such as motors or fluorescent lights.

3. How does power factor affect the apparent power in an electric circuit?

Power factor is the ratio of real power to apparent power in a circuit. A circuit with a power factor of 1 has no reactive power and therefore the apparent power is equal to the real power. However, a lower power factor means a higher amount of reactive power and therefore a higher apparent power.

4. Can real power ever be greater than apparent power?

No, real power can never be greater than apparent power in an electric circuit. This is because apparent power is a combination of both real and reactive power, so it will always be equal to or greater than the real power in the circuit.

5. How can the apparent contradiction of power in an electric circuit be resolved?

The apparent contradiction can be resolved by understanding the difference between real and apparent power, as well as the role of reactive power and power factor in a circuit. By properly measuring and managing these components, the apparent contradiction can be explained and accounted for in an electric circuit.

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