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Veritasium "How Electricity Actually Works"
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electrodacus:

--- Quote from: dunkemhigh on May 21, 2022, 09:55:34 pm ---
--- Quote ---Can you understand the relation and also the difference ?
--- End quote ---

Can you not understand that to charge the capacitor you need a completed circuit for SOMETHING to go around. If you leave off the return path then nothing will happen. Therefore something must be going down that return path.

Are you saying that this is wrong? That you don't need a completed circuit for the capacitor to charge?

--- End quote ---

Yes that is right. There is no complete circuit that is why energy stops flowing once the capacitor is fully charged.
The circuit is one source supplying a energy storage with limited capacity so current will flow from the source through the wires in the energy storage device (capacitor). No energy flows through the capacitor at any point in time.   
PlainName:

--- Quote ---so current will flow from the source through the wires in the energy storage device (capacitor)
--- End quote ---

And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.
electrodacus:

--- Quote from: dunkemhigh on May 21, 2022, 10:19:28 pm ---
--- Quote ---so current will flow from the source through the wires in the energy storage device (capacitor)
--- End quote ---

And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.

--- End quote ---

See the problem I asked snarkysparky to solve in post 784. Result form that will blow your minds.

I asked to reverse the polarity of the voltage source after the capacitor has charged so circuit is in steady state.

I can provide the answer but likely you won't believe if you do not do that yourself.
PlainName:

--- Quote from: electrodacus on May 21, 2022, 10:33:04 pm ---
--- Quote from: dunkemhigh on May 21, 2022, 10:19:28 pm ---
--- Quote ---so current will flow from the source through the wires in the energy storage device (capacitor)
--- End quote ---

And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.

--- End quote ---

See the problem I asked snarkysparky to solve in post 784. Result form that will blow your minds.

I asked to reverse the polarity of the voltage source after the capacitor has charged so circuit is in steady state.

I can provide the answer but likely you won't believe if you do not do that yourself.

--- End quote ---

I am wrong there - got carried away. But there is A current going through that resistor isn't there? Go on, you have to admit to that. Let me remind you with the simulation:

electrodacus:

--- Quote from: dunkemhigh on May 21, 2022, 10:42:14 pm ---I am wrong there - got carried away. But there is A current going through that resistor isn't there? Go on, you have to admit to that. Let me remind you with the simulation:

--- End quote ---

Of course there will be a current through the resistor as you are charging the capacitor.  Energy needs to travel through wire and the resistor is a wire.
But the new question is what happens if after you charged the capacitor so no current but capacitor contains those 72mJ and you reverse the polarity on the source.
How much energy will be delivered by the source and how much of that will end up in the resistor ?
I'm fairly certain you will be surprised by the result as it will be very different from initial result because the capacitor is already charged.

The problem data is a 12V ideal voltage source 0.5Ohm resistance for each wire so 1Ohm in total and 1000uF capacitor.
Initial connection when capacitor was discharged is 144mJ supplied by the ideal voltage source = 72mJ dissipated on wires as heat  and 72mJ stored in the 1000uF capacitor.
Then after this is done just remove the supply and connect it again with reverse polarity.
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