# Power in electric circuit: apparent contradiction

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1. Sep 24, 2014

### Ott Rovgeisha

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!

2. Sep 24, 2014

### 256bits

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. Sep 24, 2014

### Ott Rovgeisha

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. Sep 24, 2014

### 256bits

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. Sep 24, 2014

### willem2

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. Sep 24, 2014

### Ott Rovgeisha

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. Sep 24, 2014

### Ott Rovgeisha

8. Sep 24, 2014

### Drakkith

Staff Emeritus
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. Sep 24, 2014

### 256bits

Look again closely at the calculation. The new resistance for the thicker wire is R/3.

10. Sep 24, 2014

### Ott Rovgeisha

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. Sep 24, 2014

### Ott Rovgeisha

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. Sep 24, 2014

### 256bits

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. Sep 24, 2014

### Drakkith

Staff Emeritus
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. Sep 24, 2014

### Ott Rovgeisha

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!