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If:

I = V / R

And a light bulb needs X amount of electrons flowing per second (I=amps) to create light. Why do we use watts on the first place if all the light bulb cares is the number of electrons flowing per second (amps) to do the job?

My point is that, what creates the power (a.k.a. watts) for the light bulb to light up is the flow of electrons per second (amps), period. Thus watts should be the same as amps(P = I) if we just want to use another word. For example we can have a circuit with different volt and different resistance and deliver the same number of amps and both will light up the light bulb.

I = V / R 
I = 10v / 2 = 5amp 
I = 5v  / 1 = 5amp

PS: I'm talking exclusively about electricity.

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Power is always the product of an effort variable (like voltage) times a flow variable (like current): P = (e)x(f) or in this case, P = (V)x(I). in mechanical terms, we have P = (torque) x (rpm), for jet engines P = (thrust) x (velocity), and so on.

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  • $\begingroup$ Probably best answer, at the same time the simplest. $\endgroup$ – Gabriel Rodriguez Dec 25 '17 at 23:16
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My point is that, what creates the work/power(a.k.a. watts) for the light bulb to light up is the flow of electrons per second(amps), period. Thus watts should be the same as amps(P = I) if we just want to use another word.

No, you are wrong. What creates the power of the incandescent bulb is the dissipation of electrical energy at the resistance of the tungsten filament. Hence, the power dissipated depends both on the current and the resistance.

For example we can have a circuit with different volt and different resistance and deliver the same number of amps and both will light up the light bulb.

No, for similar reasons. Power $P$ is given by: $$P=I^2R=\dfrac{V^2}{R}$$

You can check wikipedia for more info.

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    $\begingroup$ Just to add the other implied but perhaps "missing" piece of information; you have near-constant voltage running through your house circuits as well, along with the set resistance of the filament. $\endgroup$ – JMac Dec 25 '17 at 17:01
  • $\begingroup$ @JMac Was going to edit it in. Thanks for the add-in. $\endgroup$ – SchrodingersCat Dec 25 '17 at 17:02
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    $\begingroup$ I suppose an "intuitive" argument that the resistance along with the current is important (caution: hand-waving ahead) is that in some sense current quantifies the energy we're pumping in, and resistance quantifies the strength of interaction of the material with that energy. Zero resistance means however much current we send through, the bulb just doesn't interact with it and doesn't light up. Could there be a way to flesh out my argument further so that it is better grounded? $\endgroup$ – Styg Dec 25 '17 at 17:15
  • $\begingroup$ @JMac so you are saying that having a 10 Volt Battery over a 2 Ohms resistance conductor wont have exactly the same effect on the light bulb as a 5 Volt Battery over a 1 Ohms resistance conductor??? (That defies all the logic and everything I have learned about current). $\endgroup$ – Gabriel Rodriguez Dec 25 '17 at 17:48
  • $\begingroup$ @GabrielRodriguez Those would be completely different lights. Current doesn't completely define the power that a light dissipates. Power is the energy per unit time that the light is using. As shown in this answer, power is $I^2 R$ or $\frac {V^2}R$. It depends on both current and resistance, so a $5V \ 1 \Omega$ light is far less powerful than a $10V \ 2 \Omega$ light. Every light has a fixed resistance. Typical household light circuits also have a (approximately) fixed voltage; so every light can be rated for the power across it. Current alone doesn't define light power. $\endgroup$ – JMac Dec 25 '17 at 17:56
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Why do we use watts on the first place if all the light bulb cares is the number of electrons flowing per second(amps) to do the job?

First, your formula $I=V/R$ only applies to incandescent bulbs, which are quickly becoming obsolete.

Even if we stick to incandescent bulbs, to make a bulb that produces more or less light, it will be made with higher or lower resistance ($R$), thus causing it to consume less or more power.

Second, incandescent light bulbs aren't the only kind of load that is connected to the power grid. Motors consume a variable amount of power depending how much mechanical load they are driving. Modern electronics (including LED or CFL light bulbs) often include regulated switching power supplies that will increase the current they draw if the supply voltage dips, consuming constant power.

The power company doesn't know what kind of loads you have connected, so they charge you for energy (power over time), not for current. Of course if the grid voltage stays constant, it amounts to the same thing.

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  • $\begingroup$ I don't care what you are putting in the circuit, the light bulb was just an example, of something requiring X amount of amps. $\endgroup$ – Gabriel Rodriguez Dec 25 '17 at 17:39
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    $\begingroup$ @GabrielRodriguez, I can only answer the question you write down, not read your mind and answer the different question that's in your head. $\endgroup$ – The Photon Dec 25 '17 at 18:18
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Two light bulbs with different filament thread materials might require the same amount of energy per second (that is the power) but achieved with different currents.

A larger material resistance will cause the same current to deliver more energy. So power and current are different things.

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  • $\begingroup$ When you say "larger material resistance" you mean i.e. thicker wire? And if so, when you change the resistance you are automatically changing the current because they are directly related anywas (I = V/R). So, how can you say "same current"??? ....The current is going to change automatically... $\endgroup$ – Gabriel Rodriguez Dec 25 '17 at 17:56
  • $\begingroup$ @GabrielRodriguez What if you change the voltage as well? What if you have a non-ohmic component? Power and current do not only follow each other through resistance; other things interfere as well. And in any case, just because properties are related, they are not the same. We say "power" because we need energy for our lamp. Easy concept to grasp. That this then happens to fit another property, namely current, is just a coincidence. $\endgroup$ – Steeven Dec 25 '17 at 21:54

protected by Qmechanic Dec 25 '17 at 18:05

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