Circuit completion : is it necessary?

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A closed circuit is essential for current to flow, which is why a bulb does not light when only one terminal is connected to a battery. Even if one terminal is attached, there is no potential difference across the bulb, preventing electron movement and thus no illumination. While static electricity could theoretically cause a brief flicker if a negatively charged body is used, this scenario still requires a complete circuit for sustained current flow. Connecting one terminal of a battery to the earth does not drain the battery, as current cannot flow without a complete circuit between both terminals. Ultimately, without a closed loop, no current can pass through the bulb, confirming the necessity of circuit completion for functionality.
  • #31
How is connecting a battery to a bulb not considered DC?

Not sure how you can have any capacitance between a battery post and a light bulb, but I suppose it could be a possiblity...
 
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  • #32
Dmytry said:
He's not discussing DC though... he's asking what will happen when he connects the battery to the bulb, which is not a steady state, and he is absolutely right that the current will flow until potentials equalize, and it really is the case that neon bulb will lit up for a short time when he connects it. You can say in this case that the circuit is completed by capacitance between other battery terminal and the bulb.

That is DC. And what capacitance are you referring to?
 
  • #33
Alright, so the bulb WILL glow, even if it is for an undetectable period of time. Well since the bulb can glow, we can also get a shock, by just touching a high voltage source, without completing the circuit.

Right?
 
  • #34
No. Bulb won't glow. You won't get a shock if only one post of the battery is involved.
 
  • #35
Could we please get at one correct conclusion? someone said the bulb will glow.
 
  • #36
When I saw the title of this thread I knew trouble was brewing.

And as always we have all sorts of red herrings introduced.

The simple fact is that if you move even one single unit of charge from a to b you have, by definition a current.

The fifth picture in this article shows a classic where a man's hair is standing on end because of proximity to a Van De Graaff generator.

http://en.wikipedia.org/wiki/Van_de_Graaff_generator

Capacitance, inductance, resonance or any other ants is not involved.

Some charge passes from the generator ball to the man by traveling through the resistive path formed in the air between him and the generator ball.

This is by definition a current.

No circuit is involved or completed.

Because the man is insulated the charge does not proceed further and spreads to his extremities, causing his hairs to separate.

If you bring your knuckles close to the ball you can actually feel the small charges jumping the gap and impinging upon your flesh.
 
  • #37
AlchemistK said:
Could we please get at one correct conclusion? someone said the bulb will glow.

I don't see how the bulb can glow if there isn't a difference in potential between the two terminals of it. Which, as far as I know, there shouldn't be if you only connect one side of a battery to it.
 
  • #38
The light will not light if we're talking a battery and a light bulb.

That is the one correct conclusion. The other conclusion is incorrect, by definition.

:smile:
 
  • #39
Don't think of the source of charge as a battery, think of it as a simply charged body.

Suppose we remove the variable of Earth and everything else and conduct the experiment in ideal situations, would your touching a current carrying wire (remember your not touching anything else) make you experience a shock?
 
  • #40
Well jumping from a battery scenario to a static electricity scenario completely changes the entire conversation.

You don't typically talk about circuits with static electricity. You're more likely to be talking about discharge then. Studiot just hit on that pretty nicely
 
  • #41
Yeah, but one more thing, don't people say that birds don't get killed while perching on live wires because they don't complete the circuit? Shouldn't charge flow into the birds ?
 
  • #42
AlchemistK said:
Don't think of the source of charge as a battery, think of it as a simply charged body.

Suppose we remove the variable of Earth and everything else and conduct the experiment in ideal situations, would your touching a current carrying wire (remember your not touching anything else) make you experience a shock?

Not if the electrons had nowhere to go to and the conductor was electrically neutral. (Which a current carrying conductor is)
 
  • #43
AlchemistK said:
Yeah, but one more thing, don't people say that birds don't get killed while perching on live wires because they don't complete the circuit? Shouldn't charge flow into the birds ?

The resistance of a bird is much much higher than the conductor itself. There really isn't any reason the current would want to flow through the bird just to get back to the same conductor when it could simply keep on moving through the conductor.
 
  • #44
It wouldn't be that all the current would just pass by, even if it has a greater resistance, there would still be some current flowing though the bird.
 
  • #45
There is no potential between the bird's first leg and his second leg. No current flows. I suppose in theory you might get some nano amps through the bird but its insignificant.
 
  • #46
Why is there not a potential difference? the wire is positively charged and the bird is neutral. Voila!
 
  • #47
AlchemistK said:
Why is there not a potential difference? the wire is positively charged and the bird is neutral. Voila!

The wire is not positively charged.
 
  • #48
sorry i meant negatively. My bad. Slip of...keys.
 
  • #49
Its not negatively charged either
 
  • #50
@.@ its not? but its got electrons flowing through it.
 
  • #51
There is no "charge" on the wire at all... Current is flowing through that wire and returning on the neutral wire (usually several feet below the "hot" wire or wires).

There is no "charge" on either wire. There is a potential difference between the hot wire and the neutral, but they aren't "charged" in the static electricity sense. Just like there is no "charge" on the positive terminal of a battery. Just like there is no "charge" on the negative terminal of a battery.

A thundercloud overhead is "charged" and it will "discharge" to the ground when the potential gets high enough and we will see a bolt of lightning between the cloud and the ground.
 
  • #52
Wait, we say something is charged if it has more positively charged particles than negatively charged ones or vice versa. If we connect two oppositely charged bodies, by say a wire, the charge will be transferred, to make the system more uniformly charged, if not neutral.
So, actually, the negative mobile particles, electrons for example, get transferred to the wire, making it negatively charged, and then to the positive body(or even a neutral one) making the system more electrically stable.
Indicating that the wire is charged at some unit time.
 
  • #53
I can't explain it any better than http://en.wikipedia.org/wiki/Electric_charge" :

"Static electricity and electric current are two separate phenomena, both involving electric charge, and may occur simultaneously in the same object. Static electricity is a reference to the electric charge of an object and the related electrostatic discharge when two objects are brought together that are not at equilibrium. An electrostatic discharge creates a change in the charge of each of the two objects. In contrast, electric current is the flow of electric charge through an object, which produces no net loss or gain of electric charge. Although charge flows between two objects during an electrostatic discharge, time is too short for current to be maintained."

(Edit: bolded a portion of the quote for emphasis)
 
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  • #54
Alright, maybe there is no NET change in charge but what oh charge at a specific time?
 
  • #55
Drakkith said:
That is DC. And what capacitance are you referring to?
Capacitance between other lead of battery and the bulb.
The point is, it is not steady state right at the moment when you connect something, there are AC components. You can do a Fourier transform on t<0 : f(t)=0, t>=0 : f(t)=1 edit: actually not a good idea, just do Fourier on a step, when you turn something on, then turn something off.

Also, the leads of battery can be at any voltages, the battery only ensures potential difference. E.g. 1.5v battery can have leads at 0v and 1.5v or at -1.5v and 0v, or at 100v and 101.5v . If you just let battery sit with equal leakage from both terminals to ground it will be at +0.75v and -0.75v but it will take a while in practice and you are unlikely to have equal leakage. That is why when you connect positive terminal of one battery to negative terminal of another battery through lightbulb (and leave the other terminals disconnected), there won't be continuous current (but there may be transient when you connect due to parasitic capacitances). That's why you can connect batteries in series and get more voltage.

The OP's question is not because he doesn't know current won't flow continuously, it's because he knows that some charge has to be moved, and he is totally correct in that. You guys however really just confuse him. The charge is very small though.
 
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  • #56
AlchemistK said:
Yeah, but one more thing, don't people say that birds don't get killed while perching on live wires because they don't complete the circuit? Shouldn't charge flow into the birds ?
There's already a completed circuit on live wires, assuming that something(s) at the end of the transmission lines is using the electricity supplied by the wires. The birds provide a high resistance and very short parallel path to the wire and only receive a tiny amount of currrent. It is possible to draw current from such wires with a transformer and/or antenna like setup (it's also illegal), and there's enough strength in the electrical field from those high voltage live wires to light up florescent bulbs stuck in the ground.

http://gadgets.boingboing.net/2009/04/10/uk-farm-grows-1301-f.html

batteries
The amount of charge on the terminals of a battery is very small, and that tiny amoung charge quickly dissipates if you connect the terminal to a large plate or Earth ground without completing the circuit. If some large conducting plate were altnerately and rapidly switched between the two terminals, eventually the battery could be drained, but it would take a very long time. Unlike a capacitor which can have a high amount of charge on it's plates a battery has a low amount of charge on its terminals relies on an inernal chemical reaction to produce current, which requires circuit completion.

battery draining in car
The primary source of leakage is probably the rectifier and/or any capacitors in the alternator used to charge the battery when the engine is running. I'm not sure this is an issue with modern cars.

circuit completion
Current can flow if you have a device that can generate a charge at one end, and another device that can drain that charge at the other end. Unless this is done in a vacuum like outer space, eventually the charge will flow back through the Earth and/or air, so there is eventual circuit completion, but it's not required to produce the original current flow.
 
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  • #57
Dmytry said:
Capacitance between other lead of battery and the bulb.
The point is, it is not steady state right at the moment when you connect something, there are AC components. .

Why is there a capacitance between a light bulb and a wire? This is something new to me... Never heard of it.

Dmytry said:
Also, the leads of battery can be at any voltages, the battery only ensures potential difference. E.g. 1.5v battery can have leads at 0v and 1.5v or at -1.5v and 0v, or at 100v and 101.5v .

Again... I'm not real sure what you're talking about. A voltage is a measurement between two points so if you're saying 1.5 Volts between the terminals... Where is the 100V measurement coming from? one of the battery teminals and what else? The ground? I don't mean to repeat myself, but if you put one meter lead on either of the posts of that 1.5V battery, you will not read a voltage until you put that other lead on the other post of the battery. if you put that lead anywhere else your meter will read 0 volts.

Dmytry said:
If you just let battery sit with equal leakage from both terminals to ground it will be at +0.75v and -0.75v but it will take a while in practice and you are unlikely to have equal leakage.

Where are these numbers coming from? What is this "leakage" you are talking about and what are these 0.75v measurements? leakage rates? Please explain...


Dmytry said:
That is why when you connect positive terminal of one battery to negative terminal of another battery through lightbulb (and leave the other terminals disconnected)

what other terminals? There are two terminals on a battery and two terminals on a light bulb... if you connected this up as stated, which terminals are disconnected?

Dmytry said:
The OP's question is not because he doesn't know current won't flow continuously, it's because he knows that some charge has to be moved, and he is totally correct in that. You guys however really just confuse him. The charge is very small though.

Saying that a charge will move through a light bulb when you connect only one lead of a battery to it does not make it so... Your explanation is that a wire and a lightbulb somehow make up a capacitor that can charge itself, but I'm not sure this is an accurate statement.

rcgldr said:
The amount of charge on the terminals of a battery is very small, and that tiny amoung charge quickly dissipates if you connect the terminal to a large plate or Earth ground without completing the circuit. If some large conducting plate were altnerately and rapidly switched between the two terminals, eventually the battery could be drained, but it would take a very long time. Unlike a capacitor which can have a high amount of charge on it's plates a battery has a low amount of charge on its terminals relies on an inernal chemical reaction to produce current, which requires circuit completion.

This is an interesting statement. Do you have a good source for information on this? I have spent considerable time in the past looking for information about charge buildup from voltage sources and I haven't been able to find a thing...
 
  • #58
rcgldr said:
The amount of charge on the terminals of a battery is very small, and that tiny amoung charge quickly dissipates if you connect the terminal to a large plate or Earth ground without completing the circuit.

Evil Bunny said:
This is an interesting statement. Do you have a good source for information on this? I have spent considerable time in the past looking for information about charge buildup from voltage sources and I haven't been able to find a thing.

Take a look at post #21 in this thread:

https://www.physicsforums.com/showthread.php?p=3116670&postcount=21
 
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  • #59
rcgldr said:
Take a look at post #21 in this thread:

https://www.physicsforums.com/showthread.php?p=3116670&postcount=21

That's what I thought, so if you hook up a light bulb to just one terminal, some charge will flow to the light bulb, and on the other battery terminal the charge just stays on so that the potential between that end of the battery and the bulb terminal is equal to the voltage of the battery. But will this cause the bulb to flash when it just receives current from one terminal?

Also, if you connect a voltmeter to one end of the battery, and the other end of the voltmeter to the ground, then charge should flow from the battery to the ground and a voltage should be recorded in the voltmeter. But this would imply that a 12 V battery should register 6V when connecting just one terminal to the ground, which doesn't seem correct. Is it because the battery does not replenish the charge on that terminal until it can accept the same amount of charge on the other terminal, so that there is not enough charge for a sustained current? So initially the voltmeter would record 6 volts but will quickly drop to zero as the terminal runs out of charge?
 
  • #60
RedX said:
Also, if you connect a voltmeter to one end of the battery, and the other end of the voltmeter to the ground, then charge should flow from the battery to the ground and a voltage should be recorded in the voltmeter. But this would imply that a 12 V battery should register 6V when connecting just one terminal to the ground, which doesn't seem correct.
It isn't. As mentioned in that post, the effective charge is miniscule, less than 10-11 Coulombs, and so will be the voltage.
 

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