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If the positive plate on a charged capacitor will pass current to the negative... |
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| Dec11-12, 05:34 PM | #1 |
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If the positive plate on a charged capacitor will pass current to the negative...
...plate of a different capacitor, why wont the positive end of a battery (lets say AA) pass current to the negative plate of a different AA battery. Does it have something to do with the chemical reaction that happens inside the battery?
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| Dec11-12, 06:01 PM | #2 |
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The plates are separated enough and the electrons cant jump from one plate to the next hence no effective current flow. Capacitors hold charge.
Here's some more info on capacitors: http://en.wikipedia.org/wiki/Capacitor |
| Dec11-12, 08:35 PM | #3 |
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lundyjb,
jedishrfu, Ratch |
| Dec11-12, 10:57 PM | #4 |
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If the positive plate on a charged capacitor will pass current to the negative...
i understood the question, actually there would be current from one battery to the other but only for a short time, because there is no continuity (circuit not formed) between them, the electrons would pile up in the "receiving battery", if you want a continue current to exist you must set a path back from the receiving battery to the first one (both would be connected in series).
in an extreme scenario you could imagine that the receiving battery would take all electrons from the first battery (there's a finite amount of electrons), because the chemical reaction will always maintain each poles at a constant potencial level, but once electrons pile up in the receiving battery they will create a strong electric field that would reject and stop the incomming electrons. |
| Dec11-12, 11:00 PM | #5 |
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Thanks for your corrections, Ratch. You are correct, although it is common to say current flow.
With the charge comment, I was really referring to the one plate being more positive (lack of electrons) and the other being more negative (excess of electrons) but you are correct the net charge is zero for the capacitor. |
| Dec12-12, 06:35 AM | #6 |
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Nevertamed,
Wrong with respect to both a battery and capacitor. According to your reasoning, a bird would not be able to roost on a high voltage wire because of a supposed short transient current it would receive when it first touched the wire. It just doesn't happen that way. A battery does not accumulate electrons on either pole, and a capacitor does not imbalance the charge on its plates unless there is a voltage difference between them. Just connecting one plate of a capacitor to a voltage source won't do anything, because there would be no voltage difference between the plates. Ratch |
| Dec12-12, 12:27 PM | #7 |
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I don't know how they know to avoid them. Perhaps the local field strength causes them to tingle and so they keep clear of such attractive perches. |
| Dec12-12, 12:46 PM | #8 |
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sophiecentaur,
Ratch |
| Dec12-12, 01:00 PM | #9 |
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| Dec12-12, 01:10 PM | #10 |
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| Dec12-12, 08:54 PM | #11 |
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Sophie is correct.
Birds tend to avoid power lines > about 30KV. I'd guess it tickles their feet. It'd be the minute capacitive current that charges their body alternately to +/- line peak voltage at line frequency , 50 or 60 hz. When working such lines hot, linemen wear a special suit that forms a Faraday cage around them. http://www.hubbellpowersystems.com/l...ories/suit.asp |
| Dec12-12, 09:11 PM | #12 |
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| Dec12-12, 09:17 PM | #13 |
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| Dec13-12, 03:08 AM | #14 |
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| Dec13-12, 04:12 AM | #15 |
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Nevertamed,
sophiecentaur, Ratch |
| Dec13-12, 05:25 AM | #16 |
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Explanations in terms of equivalent lumped components are often helpful - antenna theory is often usefully approached this way. btw If you want to be fussy about nomenclature, then I think it would be more desctiptive to say that Capacitors become Polarised, if you don't like 'charged'. (I have not read of the term "Energised" in this context - it is used more int the context of Batteries, I think but I don't think it is defined very rigorously, though people do talk of 'energising a coil' when you switch on the current, I suppose). That just means that there is a displacement (imbalance) of net charge from one side to the other. With no dielectric, the charge needs to be 'taken off' one side and 'put into' the other. When there is a dielectric, the additional charge (giving it higher Capacitance) is due to charges being easily displaced within the dielectric material as its molecules become polarised. You need to move more charges in this case for a given Voltage. |
| Dec13-12, 06:20 AM | #17 |
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we have the first cap: with one plate full of electrons and the other one lacking (after being charged or energized if you like it) we have a second cap: with one plate full of electrons and the other one lacking (after being charged or energized) if you set a path (an hypothetical wire) from the first capacitor's full of electrons plate to the second's capacitor lack of electrons plate; there will be a short current until this imbalance disappear (there wuld be a voltage between them, despite the fact the plates involved belong to different caps) about your second quote: In a way, a capacitor is a little like a battery. Although i know they work in completely different ways, capacitors and batteries both store electrical energy. A battery has two terminals. Inside the battery, chemical reactions produce electrons on one terminal and absorb electrons on the other terminal, it maintains this imbalance of charge ( charge separation) as the acid keep working in order to offer a constant voltage |
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