Circuit completion : is it necessary?

  1. It is known that a voltage must be applied around a closed circuit to make current pass through it. So thats why a bulb doesn't glow without connecting both terminals, right?

    But, suppose only one terminal of the bulb is attached to a battery and the circuit is closed, the battery still has a higher charge relative to the bulb, so electrons must still go to the bulb, to even out the charge, so why doesn't the bulb glow,even if it is for a small time period?....or does it?
  2. jcsd
  3. That's a contradiction. If only one terminal is connected, the circuit is open, by definition.

    No it doesn't. There is no place for the current to go if the ciruit is open. The electrons are already in the bulb. something needs to move them. If no potential difference is applied to the bulb then the electrons can't move. If only one terminal is connected to the bulb, you have not applied a potential difference to it.
  4. DaleSpam

    Staff: Mentor

    AlchemistK, I think you may have a misconception about how batteries work. Batteries are electrically neutral, they do not have a "higher charge relative to the bulb". What they have is a potential difference between the terminals. If you touch only one end of the bulb to one of the battery terminals then you have not put the potential difference across the bulb, so no current.
  5. Alright...i see.

    So the bulb should light in the above case if instead of a battery, a negatively charged body is used?
  6. Now you're talking about static electricity... Not sure if enough "equalizing" current would flow to make the bulb flicker. I'm thinking probably not.
  7. DaleSpam

    Staff: Mentor

    Assuming that you had a huge amount of charge you might be able to get enough current to make it flicker for a brief moment. Or you could arrange it so that it would arc from the other end of the bulb, in which case you could definitely get a flash, but that is completing the circuit in a sense so it is kind of cheating.
  8. You can probably see this with fluorescent or maybe neon bulbs which need very little current to flicker. But I think you could make a case for the circuit being completed "through thin air" as it were.
  9. DaveC426913

    DaveC426913 15,939
    Gold Member

    That's the ideas behind a ground, i.e. earth - which can act as an infinite sink for the other end of your circuit.
  10. Can you please explain? You're not suggesting that you can drain a battery by connecting only one of the battery terminals to the earth are you?
  11. HallsofIvy

    HallsofIvy 40,218
    Staff Emeritus
    Science Advisor

    Yes, he is, and yes, you can. It happens quite commonly. I used to have a car that would work well if you kept using it but if you let it sit for a week or two, the battery would drain- there was a short somewhere that was grounding the battery and draining it.
  12. I don't think so.

    In your example, one of the terminals of the battery (most likely the negative) was (edit: probably) fastened to the chassis framework of the car to use it as an intentional return path back to the source (the battery). It found it's way back to the battery through the metal parts of the car... It didn't drain itself into the ground through the tires, I assure you.

    You cannot drain a battery without completing a circuit from the positive terminal to the negative terminal.

    Connecting a wire from one terminal of your battery to the earth, even if we drove a 10-foot ground rod into the earth and hooked the wire to it, would not drain anything from the battery.

    (Some kind of failure of the chemical reactions inside the battery could render it useless, but that's a different conversation)
  13. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    I guess you are thinking about steady currents here, not currents and voltages that vary with time (for example AC mains supply).

    Actually you are right, some electrons do go into the buib, but this happens very fast. It would probably take only millionths of a second for the electric field to get into equlibrium and the electrons to stop moving again. You certainly wouldn't see the buib "glow", and in fact it would be very hard to measure what happened with voltages as small as a battery.

    You can get visible effects from this sort of thing in experiments with static electricity (Google it!) but the voltages involved are thousands of times higher than a typical battery, and even then the amount of power is much to small to make a light bulb glow.
  14. Isn't everything in some way or another a closed circuit? So one side of the battery could be connected to earth, the other to air, so both of those materials have their conductivities. Not sure what to use for the cross-sectional area or the length of the circuit though to use the formula for resistance: [tex]R=\rho \frac{L}{A} [/tex]. But technically there should be some current and the battery will drain.

    What if you bury a battery? If the earth is a really good conductor then should it not drain very fast?
  15. There is no current flow until there is a complete circuit. Something needs to conduct from the negative to the positive terminal.

    Current does not flow out of one terminal of the battery unless it is flowing into the other terminal of the battery. You cannot have one without the other.

    If you buried a battery in the earth, the dirt between the terminals would very likely conduct enough to form this complete circuit and the battery would drain. If the battery had enough voltage, then the air between the terminals could conduct and we would see an arc that would violently drain the battery.

    I repeat myself, but I need to say it again... We cannot drain the battery without a complete circuit. It will not happen until the loop from one terminal to the other is complete. We cannot take one wire buried into the ground and connect it to only one terminal of a battery and expect any current to flow. There is no potential difference between a single terminal and the ground. No potential difference means no current flow.
  16. I don't agree. There is no difference of potential between one terminal of a battery and a light bulb. I say nothing would happen.
  17. If you connect one end of a battery to a plate, and the other to a different plate, there will be flow of electrons until the potential difference of the plates is equal to the voltage of the battery.

    So if you connect one end of the battery to one terminal of the light bulb, there should be flow of charge from that end of the battery to that end of the terminal, and the opposite charge will build up at the other end of the battery. When the potential between that other end of the battery and the terminal of the light bulb is equal to the voltage of the battery, flow will stop.
  18. If you're talking about a capacitor, then yes, there is current flow. The opposite charges attract each other through the dielectric and the capacitor charges. The negative terminal releases electrons into the plate and the positive terminal accepts electrons from the other plate. electrons left the negative terminal and returned on the positive terminal. A complete circuit was made.

    If you're just talking about a couple random metal plates on the ground, I don't think much of anything happened because there were not charges to attract them to each other... there was no circuit made so little or nothing flowed into or out of the plates.

    No. Nothing moves until the other end of the battery is also in contact with the other side of the bulb. There will be no flow of charge from one end of the battery to the terminal of the light bulb because there is no difference of potential. The positive terminal is neutral to the light bulb. Just like the negative terminal is neutral to the light bulb. Neither of the battery posts are attracted to the light bulb on their own.

    The only thing the positive post of that battery is attracted to is the negative post of the same battery. It's not attracted to the ground. It's not attracted to any other battery. It's not attracted to the frame of a car, and it's not attracted to a light bulb. The only thing in the universe that positive post is attracted to is it's own negative post. That's it... nothing else.
  19. DaveC426913

    DaveC426913 15,939
    Gold Member

    So you're suggesting I cannot get a shock by grabbing the positive terminal of a car battery while standing barefoot in a puddle of water.
  20. I shall humbly offer my layman's understanding of the issue. Your argument, if I understand correctly, is that a stream of electrons (in or out), from battery-terminal potential-equalizing, should give light to the bulb for at least a short while, when the bulb is connected by one terminal to one from the battery.

    I would compare the bulb to a hydraulic engine. If you open for pressurized liquid to enter engine, while the exit flow-valve of engine is closed, you would get pressure in the engine, but no work would be done, as that depends of flow. There would be some compression of the liquid though, and this may allow some insignificant movement of the engine-rotor, as the pressures equalize through the engine.

    How eletrical charge compresses/expands in a lightbulb glow-wire is not known to me, but low volt DC will only make one, small adjustment of electron-numbers. A high volt ac-terminal would make the electrons compress and expand repeatedly, and faster, thus maybe make a glow of duration in a sensible (low-volt) bulb.

    Magnetically induced AC in short circuit glow-wire will however make a bulb glow, entirely without terminals. Much better (and safer) party-trick. Maybe shortcircuit a normal bulb and lay it on an induction cooker-plate. General warnings, of course
  21. Leaking from a cars battery occurs not because (-) is in the body of the car, but because (+) leaks to body, unintentionally.

    And if you get a shock, negative pole would not be carried through the water and bare feet, but the body of the car. The car is insulated from the ground, normally, with rubber wheels.
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