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| Two capacitor problem (open circuit version) |
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| MrAl:
--- Quote from: wraper on August 06, 2019, 12:54:29 pm --- --- Quote from: MrAl on August 06, 2019, 08:54:14 am --- --- Quote from: wraper on August 05, 2019, 02:34:58 pm --- --- Quote from: MrAl on August 05, 2019, 01:31:49 pm ---This open circuit is not good for anything until it is used for something, and then things change. --- End quote --- This is not real circuit but theoretical question. --- Quote ---Granted though this is not really a circuit yet unless we consider moisture, which will conduct very little but still some. If one cap has 10v and the other 0v, then the initial voltage for C1 is 10 and for C2 is 0. --- End quote --- These assumptions are useless. Because real capacitor has leakage. If leakage is high, result will be completely opposite to what you said when high impedance meter is connected. Also capacitors with very low leakage may charge themselves just by air movement or electric field in their vicinity. --- End quote --- Useless to you not to me. --- End quote --- I mean don't add things which do not exist in the question :palm:. Because if you are starting adding things like meter, it becomes different circuit. Suddenly there are variables which depend on implementation. What if practical implementation is capacitor used as voltage storage (memory) which is disconnected by relay. It stays under voltage it was charged with, no load connected. If there is capacitor in the thing, it does not mean there is any load connected which is discharging it. Also if capacitor is big enough, discharge by multimeter will be insignificant. The only thing you are doing is confusing OP. --- End quote --- It's not up to you whether or not i add things. It's up to me. But the voltage across capacitors at t=0 is always called the initial voltage. It does not depend on whether or not you discharge it or not because the implication is that some day it will be discharged (or charged more). Look into any circuit simulator out there and see how they start capacitor voltages at t=0. You have your choice to allow the default 0v or some other set voltage you choose. Now a constant voltage source is different because it has a set voltage that does not change. The cap is not equivalent to a voltage source it's a different kind of component therefore it has different properties. So the only thing you are doing is confusing the OP. |
| wraper:
--- Quote from: MrAl on August 07, 2019, 03:36:40 am ---Look into any circuit simulator out there and see how they start capacitor voltages at t=0 --- End quote --- Look into any circuit simulator and watch how voltage remains constant with time passing. I already said that: --- Quote from: MrAl on August 05, 2019, 05:58:30 am ---You cant really say that the voltage at point "B" is 1000v, you can only say that the *initial* voltage at point B is 1000v. That's because once you use this circuit for something, that second cap will start to charge. Even if you go to measure that voltage with a meter the cap will start to charge a little. --- End quote --- is pointless because --- Quote from: wraper on August 05, 2019, 11:48:34 am ---And you can say "initial" just about anything. Like when connecting 2 capacitors in parallel, one charged, one discharged. Initially one cap will be fully charged and another one fully discharged. Exactly the same initial result as with this not closed circuit. --- End quote --- With any circuit with these capacitors initial voltage will be the same. But what will happen later will be very different. |
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