Circuits (again)

[MedPR]

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When a capacitor discharges, does any of that charge go back into the battery? In other words, is the chemical reaction within the battery initially a reversible reaction? I'm assuming not..?

Also, theoretically, is it possible for a battery to completely discharge into a capacitor (or group of capacitors)? If so, would the capacitors just power the circuit until all of their charge was dissipated and then, at that point, nothing would happen? There would simply be no more current (and thus no power) through the circuit?

 

[chiddler]

When a capacitor discharges, does any of that charge go back into the battery? In other words, is the chemical reaction within the battery initially a reversible reaction? I'm assuming not..?

Also, theoretically, is it possible for a battery to completely discharge into a capacitor (or group of capacitors)? If so, would the capacitors just power the circuit until all of their charge was dissipated and then, at that point, nothing would happen? There would simply be no more current (and thus no power) through the circuit?

a capacitor only discharges when the voltage source is disconnected or when the potential of the battery drops so your question doesn't make too much sense.

but if you attach a batt of lower voltage after the capacitor is fully charged, i think it should charge the battery a bit.

It should not be possible for a battery to completely discharge into a capacitor because as the battery drains, its potential is lowered until that the capacity starts producing a higher voltage. At that point, current reverses.

To help illustrate. You have a 10 V battery next to a 5 V battery, but the two positive ends are facing each other. Result? A 5 V circuit in the direction of the 10 V battery positive node.

 

[MedPR]

a capacitor only discharges when the voltage source is disconnected or when the potential of the battery drops so your question doesn't make too much sense.

but if you attach a batt of lower voltage after the capacitor is fully charged, i think it should charge the battery a bit.

It should not be possible for a battery to completely discharge into a capacitor because as the battery drains, its potential is lowered until that the capacity starts producing a higher voltage. At that point, current reverses.

To help illustrate. You have a 10 V battery next to a 5 V battery, but the two positive ends are facing each other. Result? A 5 V circuit in the direction of the 10 V battery positive node.

Thanks, I didn't know that the capacitor would only discharge when the battery was disconnected. yea, I understand the thing about two batteries and one overpowering the other to reverse the current (like in an electrolytic cell), but I can't seem to apply that to what I'm trying to figure out.

So you're saying a battery can't really fully discharge because once the voltage in the capacitor is greater than what is left in the battery, the capacitor will "take over" and reverse the current. Since the capacitor is now powering the circuit, the battery will stop discharging (stop converting chemical to electrical) so it will end up holding onto whatever energy it has left? But wouldn't the current reverse again when the capacitor discharged to a level lower than the battery? Then it would just go back and forth until both the battery and capacitor both had infinitesimal charge?

 

[chiddler]

And the circuit loses energy to whatever intrinsic resistance it has, too. Probably won't even go back and forth - would just cease when the voltages are equal. Then both lose voltage as they discharge.

When did you think a capacitor discharged?

I'm not asking to rub it in. Because i'm curious what you thought.

 

[MedPR]

And the circuit loses energy to whatever intrinsic resistance it has, too. Probably won't even go back and forth - would just cease when the voltages are equal. Then both lose voltage as they discharge.

When did you think a capacitor discharged?

I'm not asking to rub it in. Because i'm curious what you thought.

Honestly, I wasn't sure. I've never encountered an explanation nor a problem that explicitly tested whether or not you knew it. It makes sense why a capacitor would discharge when there was no other emf source, though.

However, why when the capacitor starts to discharge (because the battery has become too weak) does the current have to reverse? Couldn't the capacitor just continue pushing charge in the same direction the battery was?

Edit: Nevermind about that question. The reason it has to reverse is because current flows from the (+) end to the (-) end right? So once this battery gets too weak, the capacitor starts to discharge and pushes current in the opposite direction. So the reason why charge from a battery and from a capacitor have to go the "long way" (through the wire) around to get from (+) to (-) is because the wire has less resistance than the air between the charged plates, right? So resistors can pass charge through the "short way" (doesn't have to go all the way around) from (+) to (-) because they have a wire or some other conductive material through them instead of an insulating space (like air) that exists only in capacitors and batteries?

Edit 2: As you can see, circuits are pretty much the worst possible topic that could pop up on my MCAT. That and a complicated orgo passage.

 

[chiddler]

Think about what that implies. If a capacitor can push current in the same direction that the battery does, what would be the effect on voltage of a circuit?

Actually, draw it out. I think that'll help a lot. Specifically, draw how the current looks like when charging, and when the battery is gone and capacitor discharges.

edit: yuppers.

more edit: yes current doesn't flow from plate to plate via air. way too much R. but it is possible with a high enough voltage and, if you recall, the term is dielectric breakdown.

 

[MedPR]

Think about what that implies. If a capacitor can push current in the same direction that the battery does, what would be the effect on voltage of a circuit?

Actually, draw it out. I think that'll help a lot. Specifically, draw how the current looks like when charging, and when the battery is gone and capacitor discharges.

edit: yuppers.

more edit: yes current doesn't flow from plate to plate via air. way too much R. but it is possible with a high enough voltage and, if you recall, the term is dielectric breakdown.

What's the answer to your question? If the capacitor could push current in the same direction as the battery, wouldn't it be the same situation as if the voltage from the battery went all the way around the circuit without being dissipated -- so there would be a short circuit?

Also, can you explain what dielectric breakdown means? I know what dielectrics do and how they work, and I remember reading the term but I'm not exactly sure what happens or what the importance is. Does a low dielectric breakdown point mean that the capacitor will hold less charge than one with a high dielectric breakdown?

 

[chiddler]

What's the answer to your question? If the capacitor could push current in the same direction as the battery, wouldn't it be the same situation as if the voltage from the battery went all the way around the circuit without being dissipated -- so there would be a short circuit?

Also, can you explain what dielectric breakdown means? I know what dielectrics do and how they work, and I remember reading the term but I'm not exactly sure what happens or what the importance is. Does a low dielectric breakdown point mean that the capacitor will hold less charge than one with a high dielectric breakdown?

If it could push current in the same direction as the battery, nothing would happen because as soon as charge builds up there, it would be continue onto the circuit. It would be like having some expensive wiring. Furthermore, since its all being passed along the circuit, no charge would build up. Needless to say, this is impossible and doesn't happen anyway.

Dielectric breakdown means that the voltage stored in the capacitor is so great that current can go through the dielectric separating the capacitors. This is why lightning occurs, too. Voltages so high that current can conduct through air.

All else equal, you're right. Because a lower breakdown voltage means it has less capacity.

 

[MedPR]

If it could push current in the same direction as the battery, nothing would happen because as soon as charge builds up there, it would be continue onto the circuit. It would be like having some expensive wiring. Furthermore, since its all being passed along the circuit, no charge would build up. Needless to say, this is impossible and doesn't happen anyway.

Dielectric breakdown means that the voltage stored in the capacitor is so great that current can go through the dielectric separating the capacitors. This is why lightning occurs, too. Voltages so high that current can conduct through air.

All else equal, you're right. Because a lower breakdown voltage means it has less capacity.

Ok great, thank you!

 

[Class1P]

What if your capacitor holds a 1000 volt potential and then you hook it up to a small battery even then can the electrons push back onto the negative side of the battery?