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Circuits

So say you a typical setup like in this diagram.

  1. How does current flow while the capacitors charge? To elaborate, what is the mechanic that makes capacitors mimic the functionality of a wire instead of acting like the gap in the wire that they are?

  2. When the capacitor is fully charged, the left plate is positive and the right plate is negative right? So does that mean when you disconnect the battery the capacitor discharges in the opposite direction that the batteries current was in?

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  • \$\begingroup\$ Where did you get those images from ? They are not correct. A lamp/led will not light up continuously like that with batteries if a capacitor is placed in series. It behaves as the gap you see in the symbol for a capacitor \$\endgroup\$ – efox29 Mar 25 '15 at 23:26
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    \$\begingroup\$ I assume the intention of the graphic is to show the lamps illuminating for the brief moment before the capacitors are charged. \$\endgroup\$ – Dan Laks Mar 25 '15 at 23:28
  • \$\begingroup\$ @efox29 exactly what Dan said. I just needed an image and that was straight from my hw so it was the easiest to get. \$\endgroup\$ – m0meni Mar 25 '15 at 23:41
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  1. The mechanism is electrostatic repulsion -- the Coulomb force. The battery forces extra electrons onto the negative plate of the capacitor. The excess charge pushes an equal number of electrons off of the other plate and into the rest of the circuit. To the rest of the circuit, it looks like there's a current through the capacitor, since electrons go into one end and come out the other.

  2. If you replace the battery with a short circuit, the capacitor will indeed produce a discharge current in the opposite direction of the charging current. (It's sort of like a spring.) If you replace the battery with an open circuit, no current flows and the capacitor remains charged.

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  • \$\begingroup\$ Why causes the battery to force electrons onto the negative plate in the first place though? Why is there a voltage if there is a gap? \$\endgroup\$ – m0meni Mar 25 '15 at 23:54
  • \$\begingroup\$ There is a voltage because the battery is there. The voltage is between the + and - of the battery regardless of whether there is any connection between them. The voltage is a potential difference between the two electrodes. \$\endgroup\$ – Majenko Mar 26 '15 at 0:08
  • \$\begingroup\$ @Majenko but doesn't there need to be a connection between the two terminals of the battery for there to be a flow? How does a capacitor act like this connection if it is essentially a hole in the wire. If it was simple a gap in the wire and not a capacitor, for example, then there would never be any current. \$\endgroup\$ – m0meni Mar 26 '15 at 0:41
  • \$\begingroup\$ See my analogy below. It explains quite well how it behaves. A simple gap would be like a rubber bung in the mouth. The capacitor, because of the very large surface area, acts very very differently. And yes, a simple gap in the wire is itself a capacitor - just one with incredibly small capacitance. \$\endgroup\$ – Majenko Mar 26 '15 at 0:48
  • \$\begingroup\$ @AR7 Also, think about how the electrons actually move. They get either attracted to or repelled from one end of the battery. One half of the capacitor (say the plate attached to the + of the battery) could be like a hose with one end in water (the hose is the wire, the water tank is the capacitor plate). You can suck on the hose and pull some of the water (electrons) out of the tank leaving it with less water (electrons) in it. That lack of electrons (positive charge) attracts electrons towards the opposite plate (like a balloon to a cat). \$\endgroup\$ – Majenko Mar 26 '15 at 0:53
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The Master of Analogies is here again...

You can think of the capacitor as a balloon.

While you're blowing the balloon up air is flowing out of your lungs (battery) and into the balloon (current flowing to charge the capacitor). As the balloon increases in size it displaces the air around it (the repulsed charge from the opposite plate of the capacitor). That is, until you have blown the balloon up as far as you can, at which point the air (current) can no longer flow.

If you put too much air pressure (voltage) the balloon bursts (just like a capacitor - have you ever seen an electrolytic explode?).

If you look at the charge curve of a capacitor it's actually very similar to the air flow curve you get when blowing up a balloon - rapid initial inflation then it gradually tapers off as it gets harder to get more air into it.

And yes, when you release the balloon the air flows out the way it came - in reverse.

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    \$\begingroup\$ I spent a minute trying to twist "Majenko" into "Master of Analogies", but I don't see it. \$\endgroup\$ – Greg d'Eon Mar 25 '15 at 23:44
  • \$\begingroup\$ Good analogy here sir Majenko. +1 \$\endgroup\$ – Alexander Sabiona Mar 25 '15 at 23:55
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    \$\begingroup\$ @Gregd'Eon Don't worry, I'm twisted enough already. \$\endgroup\$ – Majenko Mar 26 '15 at 0:31

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