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Veritasium "How Electricity Actually Works"

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timenutgoblin:

--- Quote from: electrodacus on May 03, 2022, 12:40:07 am ---If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi

--- End quote ---

One other thing I want to point out is that if the final voltage on both capacitors is 70.7% of the initial voltage then you will NOT be conserving charge, but only conserving energy instead. Initially, the charge is Q = CV before the switch is closed. After the switch is closed, if the final voltage is 0.707*Vi on both capacitors then the final charge will be C*V*0.707 + C*V*0.707 = 1.414*Q where Q is the initial charge. Where does the extra 41.4% charge come from if it wasn't there initially?

rfeecs:

--- Quote from: electrodacus on May 03, 2022, 11:33:03 pm ---You are wrong about the fact that it is of topic.
There is no difference in this context between a battery and a capacitor as they are both the source of energy in the experiment.
Understanding what energy is and how capacitors work (energy storage devices) is critical to understand the problem.

--- End quote ---

There is a difference between a battery and a capacitor.
I understand how capacitors work.
Now, energy is another issue.  :-//  But I can always look up the textbook definition.
What is your definition of energy?

What did you think of the video?

hamster_nz:
The unasked question so far is:

Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe.

You balance the cross-sectional area such that the wire and pipe have the same mm^2.  After a constant current power supply is connected and the system reaches a steady state. How much current would flows through the inner wire, and how much through the outer pipe?

Three different schools of thought:

1) The pipe is a (oddly shaped) Faraday cage. There will be minimal potential difference across the inner wire, so minimal current will flow in it. The bulk of the current flows in the pipe. If connected in parallel outside of the pipe then more current would flow.

2) The current flow is based on the cross-sectional area of the conductor and its bulk resistance, so half the power will be in the inner wire, and half in the outer pipe. The same current would flow the wire was connected in parallel outside the pipe.

3) Well, it's not that simple, as the field from the pipe will still be interacting with the wire and the other way around. You would need to know about a lot of other things about the materials and geometries before you could answer this.

I'm firmly in camp 2 for steady state. But during the transient when current source is connected I'm quickly jumping into "camp 3".

T3sl4co1l:

--- Quote from: hamster_nz on May 03, 2022, 10:01:59 pm ---Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.

--- End quote ---

Don't let him distract you with added switches, or odd ratios of capacitor: again, it adds nothing to the problem, and only deflects from the central point.

Tim

electrodacus:

--- Quote from: timenutgoblin on May 03, 2022, 11:38:06 pm ---
--- Quote from: electrodacus on May 03, 2022, 12:40:07 am ---If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi

--- End quote ---

One other thing I want to point out is that if the final voltage on both capacitors is 70.7% of the initial voltage then you will NOT be conserving charge, but only conserving energy instead. Initially, the charge is Q = CV before the switch is closed. After the switch is closed, if the final voltage is 0.707*Vi on both capacitors then the final charge will be C*V*0.707 + C*V*0.707 = 1.414*Q where Q is the initial charge. Where does the extra 41.4% charge come from if it wasn't there initially?

--- End quote ---

The important part is conservation of energy not charge.
You start with 4.5Ws of stored energy in the charged capacitor 0.5 * 1F * 3V2. That is what you need to justify not charge unless you are confusing charge with energy and that seems to be the case.

Initial charge 1F * 3V = 3 Coulomb  = 0.833mAh
4.5Ws / 3600s = 1.25mWh
3V * 0.833mAh = 2.5mWh if the voltage will have been constant but voltage drops linearly so is half that 1.25mWh as calculated with the formula for Energy in a capacitor.

At the end of the experiment you can have for ideal case
2.121V on each capacitor
So 1.25mWh initial energy divided between the two capacitors so 0.625mWh in each
If you care about charge on each capacitor (not relevant) 2.121 Coulomb / 3600 = 0.589mAh
(2.121V /2) * 0.589mAh = 0.625mWh

As you see both energy and charge checks out in my calculation as I did the correct assumption that energy is conserved not charge.   

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