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Circuit completion : is it necessary?

by AlchemistK
Tags: circuit, completion
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Dmytry
#55
Apr27-11, 11:45 AM
P: 505
Quote Quote by Drakkith View Post
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.
rcgldr
#56
Apr27-11, 12:03 PM
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Quote Quote by AlchemistK View Post
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 eletrical field from those high voltage live wires to light up florescent bulbs stuck in the ground.

http://gadgets.boingboing.net/2009/0...ws-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.
Evil Bunny
#57
Apr27-11, 03:12 PM
P: 237
Quote Quote by Dmytry View Post
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.

Quote Quote by Dmytry View Post
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.

Quote Quote by Dmytry View Post
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...


Quote Quote by Dmytry View Post
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?

Quote Quote by Dmytry View Post
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.

Quote Quote by rcgldr View Post
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...
rcgldr
#58
Apr27-11, 06:49 PM
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Quote Quote by rcgldr View Post
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.
Quote Quote by Evil Bunny View Post
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:

http://www.physicsforums.com/showthr...0&postcount=21
RedX
#59
Apr27-11, 07:34 PM
P: 969
Quote Quote by rcgldr View Post
Take a look at post #21 in this thread:

http://www.physicsforums.com/showthr...0&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?
rcgldr
#60
Apr27-11, 07:49 PM
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Quote Quote by RedX View Post
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.
RedX
#61
Apr27-11, 08:26 PM
P: 969
Quote Quote by rcgldr View Post
It isn't. As mentioned in that post, the effective charge is miniscule, less than 10-11 Coulombs, and so will be the voltage.
But the voltage between the two charges on the terminals of the battery is 12 V. So if you define [tex]V_+ [/tex] and [tex]V_- [/tex] as the voltages between the positive terminal and the ground, and the negative terminal and the ground, respectively, then [tex]V_+-V_-=12 [/tex] volts, and why not assume symmetry to get [tex]V_+=V_-=6 [/tex] volts?
rcgldr
#62
Apr27-11, 09:36 PM
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Quote Quote by RedX View Post
But the voltage between the two charges on the terminals of the battery is 12 V. So if you define [tex]V_+ [/tex] and [tex]V_- [/tex] as the voltages between the positive terminal and the ground, and the negative terminal and the ground, respectively, then [tex]V_+-V_-=12 [/tex] volts, and why not assume symmetry to get [tex]V_+=V_-=6 [/tex] volts?
Although not a great analogy, think of a battery as a capacitor with a small amount of charge on each plate but with a very large distance between the plates in order to achieve that 12 volts. Now imagine a long grounded wire that spans the distance between the plates (without touching the plates). Since the charge on the plate at each end of the capacitor is small, the voltage between each plate and each end of that long grounded wire is also small.

The issue here is voltage is a potential that is affected by distance within a field, or the equivalent of distance in the case of a battery. When you measure the voltage between either terminal of a battery and some common ground, the equivalent of the distance component is much smaller between the terminal and the common ground, so the voltage is much less, even though the voltage between the terminals is relatively high.
AlchemistK
#63
Apr27-11, 11:41 PM
P: 158
I think the problem arising here is that a battery has too little voltage, so lets just take some other high voltage source into consideration.
Dmytry
#64
Apr28-11, 01:11 AM
P: 505
Quote Quote by Evil Bunny View Post
Why is there a capacitance between a light bulb and a wire? This is something new to me... Never heard of it.
never heard of parasitic capacitances? They're very small.
The thing is, OP's actually smart, he thought of parasitic capacitances himself (those are what holds the charges on battery leads etc), he just doesn't know the word.
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.
But that was electric potential I was speaking of.
http://en.wikipedia.org/wiki/Electric_potential

100 volts to what ever. Ground that is under your feet, or 'ground' that is the ground bus, or the box that the circuit was built in.
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...
suppose you have battery laying somewhere. You may have few hundred gigaohm resistance from each lead of battery to the ground, parasitic resistances.
The actual circuit would be - each end of battery connected to the ground with very high value resistor. If you use a voltmeter that has very high internal resistance (essentially infinite), you can measure the potential difference between lead and ground.
RedX
#65
Apr28-11, 01:19 AM
P: 969
Quote Quote by rcgldr View Post
Although not a great analogy, think of a battery as a capacitor with a small amount of charge on each plate but with a very large distance between the plates in order to achieve that 12 volts. Now imagine a long grounded wire that spans the distance between the plates (without touching the plates). Since the charge on the plate at each end of the capacitor is small, the voltage between each plate and each end of that long grounded wire is also small.

The issue here is voltage is a potential that is affected by distance within a field, or the equivalent of distance in the case of a battery. When you measure the voltage between either terminal of a battery and some common ground, the equivalent of the distance component is much smaller between the terminal and the common ground, so the voltage is much less, even though the voltage between the terminals is relatively high.
This just seems like it violates Kirchoff's rule that the sum of the voltages should be zero. If you have an imaginary path from one terminal of the battery to one end of the grounded wire, through the grounded wire to the other end of the grounded wire, and through that end of the grounded wire to the other terminal, then it should equal the voltage drop of the battery. The potential through the grounded wire is zero since it's a conductor. So the sum of the two voltage drops from terminal to ground wire should equal 12 V. So at the very least one of the drops has to be 6V (it could be 10 and 2V, 11 and 1V, etc, with the minimum drop 6 and 6V).

In any case if you have a single charge above a conducting plate, then the potential above the plate can be gotten by reflecting that charge behind the plate. So if you model the battery as two charges separated by a distance, then there reflections behind the surface of the earth should allow you to calculate the voltage. So for example say the plus and minus charges of the battery are at (x,y)=(-1,4) and (1,4) respectively. Then if you pretend there are also plus and minus charges at (x,y)=(-1,-4) and (1,-4) respectively, then you can calculate the potential at y>=0.

In any case Kirchoff's rule should apply.

So...

Q: How many physics students does it take to change a light bulb?

A: Not more than one, for then they can't agree on what will happen!
Dmytry
#66
Apr28-11, 01:38 AM
P: 505
Quote Quote by AlchemistK View Post
I think the problem arising here is that a battery has too little voltage, so lets just take some other high voltage source into consideration.
Well i think that'd really confuse everyone... batteries are easier and the principle is same regardless of voltage. Basically, the battery works by making the potential difference between it's leads be what's specified.
Very little charge is stored on the leads; you can say that a lot of charge is stored inside the battery chemically, but this charge is stored in a balanced way - you can not take out just the positive or just the negative charge alone.
The potentials themselves can be anything. For example, you can put battery on insulated table, and then put electrostatic charge on it (using plastic bag for example). That won't change the charges stored inside battery, you won't really charge the battery in the sense in which the battery charger does. However, this way you can make potentials on the leads be e.g. 10 000 v and 10 012 v (to ground). Then if you connect one to ground, current will flow for a short time. It's really same as if you had piece of metal in place of battery, you can put charge on it, you can get charge off it, etc. It's very confusing to put net charge onto battery lol.

You need to distinguish between two equal opposite charges 'stored' inside the battery, and the net charge of the battery.
AlchemistK
#67
Apr28-11, 01:58 AM
P: 158
Can the power supply in our house also be treated like a battery?
Then we shouldn't get a shock if we put one finger in the live wire socket until we put another finger in the neutral wire socket?
RedX
#68
Apr28-11, 02:15 AM
P: 969
Quote Quote by AlchemistK View Post
Can the power supply in our house also be treated like a battery?
Then we shouldn't get a shock if we put one finger in the live wire socket until we put another finger in the neutral wire socket?
The power supply can continuously push current through you. The battery won't push more current through you until you can get rid of the charges on the other lead of the battery.

My guess is you'll probably die if you stick your finger in the live wire socket, although I'm not completely sure since if the socket has a ground fault interrupter, wouldn't it trip?

Also, what would happen if the neutral wire weren't grounded/earthed? Then would you die if you touch the neutral wire? What if the neutral wire were only connected to the transformer and there was no connection between the neutral wire and the ground?
rcgldr
#69
Apr28-11, 02:50 AM
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Quote Quote by rcgldr View Post
Although not a great analogy, think of a battery as a capacitor with a small amount of charge on each plate but with a very large distance between the plates in order to achieve that 12 volts. Now imagine a long grounded wire that spans the distance between the plates (without touching the plates). Since the charge on the plate at each end of the capacitor is small, the voltage between each plate and each end of that long grounded wire is also small.
Quote Quote by RedX View Post
This just seems like it violates Kirchoff's rule that the sum of the voltages should be zero.
In my example, that grounded wire is never connected to either plate, but is placed much closer to the plates than the distance between the plates. Since voltage is equal to electrical intensity x distance, the much smaller distance means the voltage between either plate to the wire is much less than 1/2 the voltage from plate to plate.

If a charged particle traveled from either plate to the grounded wire, it gains much less energy than if the wire wasn't there and the charged particle traveled from one plate to the other.

Quote Quote by RedX View Post
Also, what would happen if the neutral wire weren't grounded/earthed? Then would you die if you touch the neutral wire? What if the neutral wire were only connected to the transformer and there was no connection between the neutral wire and the ground?
In most places in the USA, the neutral wire is center tapped from the last step down transformer, and it's not grounded directly, but is supposed to be within 10 volts of earth ground due to grounding at previous transformer stages. In my home I see .3 to .5 volts between "netrual" and the actual grounded 3rd pin on the 110 volt outlets in my home.
sophiecentaur
#70
Apr28-11, 03:04 AM
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It's all a matter of Capacity.
Imagine the battery and bulb connected by a large capacitor in series. You would have a series RC circuit. At switch on, you would find that current would flow (lighting the bulb a bit) until the Capacitor was charged. Leaving out the huge Capacitor gives you a minute capacitor (the gap between the two wire ends). Only a minute current will flow for a very short time.
RedX
#71
Apr28-11, 03:09 AM
P: 969
Quote Quote by rcgldr View Post
In most places in the USA, the neutral wire is center tapped from the last step down transformer, and it's not grounded directly, but is supposed to be within 10 volts of earth ground due to grounding at previous transformer stages. In my home I see .3 to .5 volts between "netrual" and the actual grounded 3rd pin on the 110 volt outlets in my home.
According to this picture at hyperphysics:

http://hyperphysics.phy-astr.gsu.edu...ele/gfault.gif

the neutral wire is grounded at the centertap, and to the same "neutral tie block" as the ground wire.

I agree that it makes no sense. We've already established that 10V is not dangerous (touching the terminals of a car battery), so why not just connect the ground wire to the neutral wire instead of driving a pin into the ground to connect to the ground wire? Can't you just connect the ground wire to the neutral wire instead?
Drakkith
#72
Apr28-11, 03:24 AM
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I agree that it makes no sense. We've already established that 10V is not dangerous (touching the terminals of a car battery), so why not just connect the ground wire to the neutral wire instead of driving a pin into the ground to connect to the ground wire? Can't you just connect the ground wire to the neutral wire instead?
The mains power lines that provide power for homes and business runs on AC electricity. The two lines that run into your house are both the positive AND negative lines at the same time. When the AC cycle is at one phase one line is + and the other -, and when the cycle reverses so do the lines. The ground is simply there for any accidents so that the current will flow to ground instead of through your toaster and into you.


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