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This is a really confusing question for me. Suppose, there is an electric circuit. And, right near to the negative terminal of the battery, I place 3 bulbs, which would take almost all the energy of the electrons(energised charge). Now, if the charge looses all of the energy, will it not stop? But if it stops, it is still under the influence of the electric field, so it should move towards the positive terminal, and because it is in the field, it still has some potential energy.

So my question:

If what I just said is right, will the charge not have infinite energy until it moves towards the positive terminal(because of the electric force)? But if this is true, shouldn't infinite bulbs light up in the circuit?

If I am wrong, then what happens to the charge when it looses its energy? What happens to the electric field?

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    $\begingroup$ The energy in an electric circuit is not in the charges but in the electric field. It's potential and not kinetic energy. The movement of charges shapes the potential drop (i.e. the electric field) along a circuit. Where charges can move more easily the potential drops less per distance (i.e. the field is smaller), where it takes more energy to move them, the potential drops more quickly (and the field is larger). Think about it for a series circuit made from multiple resistors of different resistance. $\endgroup$
    – CuriousOne
    Commented Feb 5, 2016 at 15:02
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    $\begingroup$ @CuriousOne Satisfy my wonder, if you please: why do you always post answers in the comments? $\endgroup$
    – Asher
    Commented Feb 5, 2016 at 15:12
  • $\begingroup$ @Asher: Usually because I am not sure if my answer is sufficient to make the OP understand. I admit that that's the wrong metric for the site. $\endgroup$
    – CuriousOne
    Commented Feb 5, 2016 at 15:14
  • $\begingroup$ Your tag is wrong. $\endgroup$ Commented Feb 5, 2016 at 16:14
  • $\begingroup$ @CuriousOne fair enough. I often start an answer and then discard it for the same reason. Or because I realize I don't actually know the answer... Ha ha. $\endgroup$
    – Asher
    Commented Feb 5, 2016 at 16:35

3 Answers 3

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The motion of the charges is totally unlike you setting off from home walking 10 km and feeling a little tired and slowing down and then walking another 10 km and feeling even more tired and slowing down and then walking a final 10 km and arriving home at a very slow pace, indeed just reaching your front door and stopping.

Here is a simple model of what happens.

The battery sets up an electric field inside the wires.
The mobile charge carries, the free electrons which are responsible for electrical conduction, are accelerated by this electric field and gain kinetic energy. Whilst in a bulb, any one of the three, the free electrons collide with the bound ions and give those bound ions some kinetic energy.
Then before another collision with a bound ion the free electron gains some more kinetic energy from the electric field to then give some to the next bound ion it meets.

The net effect of all this is that the free electrons have an average velocity (drift velocity) and the bound ions vibrate more, their "temperature" increases. The battery is the source of the energy which ultimately resides with the bound ions. The free electrons are just the mechanism by which the energy is transferred. Even if a free electron lost all of its kinetic energy it can then take some more from the electric field so it will never be marooned.

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  • $\begingroup$ I understood all of what you just said, plz also answer my second question of infinite energy and bulbs $\endgroup$
    – codetalker
    Commented Feb 5, 2016 at 15:24
  • $\begingroup$ It will not have energy because it will take time for the free electron to accelerate and then it will suffer another collision. It is a dynamics process which the free electrons being the carriers of energy between the battery and the bound ions in the light bulb. If I have not answered your second question then please will you rephrase it? $\endgroup$
    – Farcher
    Commented Feb 5, 2016 at 15:37
  • $\begingroup$ I wanted to ask this: Suppose the 3 bulbs take all the energy of the charge. Now, under the influence of the electric field, the charge again gains some energy, and moves again. So, after the 3 bulbs are lit, will the charge, after passing through the 3 bulbs, light another bulb, and another bulb thereafter, and so on? $\endgroup$
    – codetalker
    Commented Feb 5, 2016 at 15:57
  • $\begingroup$ Yes the charge can light as many bulbs as you like but with extra bulbs the E-field strength will be lower so between collisions the free electrons will not gain as much kinetic energy and so the bound ions will not vibrate as much as before. So the filaments will not be as bright. This is common experience if you connect bulbs in series. $\endgroup$
    – Farcher
    Commented Feb 5, 2016 at 16:10
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  1. Assuming incandescent bulbs rather than, say, LEDs: they won't take "almost all" the energy, they'll take all of it. It doesn't matter if you use one bulb or five, it'll use up all of the energy that the battery is supplying. That's why multiple bulbs will be dimmer than one single bulb.

  2. Electrons collectively don't have a very good sense of how much energy they have. In a good conductor even the slightest electric field causes a ton of flow of electrons to compensate; so we generally assume that the "wires" have a constant voltage all the way across. If you like, you can pretend that what the electrons are doing is something like the way continuity works in a system of pipes of water: if you pull some water out from this side of this pipe into this machine, it must therefore pull the rest of the column of water in the pipe forward as well; Nature's not going to just let you create a vacuum there.

  3. No, there is no "infinite energy." A battery takes electrons from the positive side of the terminal and absorbs them into itself, putting new electrons with a higher energy on the negative side of the terminal. If it's not connected to anything then it will do this until the net charge across the battery cancels out its own tendency to push electrons from $+$ to $-$. This does not happen at an infinite charge density or an infinite voltage; if it did then every battery not in a vacuum would be constantly discharging via sparking through the air.

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  • $\begingroup$ I was hoping for a different explanation... I understood whatever you said, but the thing I am still not able to understand is what happens to the charge when the load takes all its energy.... Does it stop? But, it is still under the electric field, which means it still has some potential energy... and can thus do more work in the circuit. Then why do we say that the charges loose energy? $\endgroup$
    – codetalker
    Commented Feb 5, 2016 at 15:09
  • $\begingroup$ @Siddhantinf: I can't give you a different explanation, only a correct one. The charge does not "stop". The current keeps flowing because you're drawing current to go through the bulbs and that current has to come from somewhere. $\endgroup$
    – CR Drost
    Commented Feb 5, 2016 at 15:25
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Lets take your language.Charges lose their energy as they move in resistor. But they don't lose all their energy and have enough energy to come back to positive terminal. Even if they lose all their energy the electrons are attracted by positive terminal and gain energy again.

If you will apply infinite bulbs their resistance would be so high that no current flow in circuit. None of your bulb glows.

Now connect them in parallel.

Now, as electron begin to move in circuit all their energy is finished in seconds and battery is empty.

Now, move into wire.

When your electron enters bulb at that very moment other electron from other side leave bulb. Electron which enter now lose its KE to glow bulb. Electron which left wire now glows another bulb if in series. If it were in parallel it would return to battery.

Now, electron connected in series you would think can light infinite bulbs.

Now,your answer comes-

If you will connect so many bulbs, length of wire would increase, hence , electric field in wire decrease.

So much reduced field cannot accelerate many electrons. Less charge leaves circuit and bulbs glow very dim.

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