Electronics > Beginners
Two capacitor problem (open circuit version)
SparkyFX:
I think answers might be different when this is either about ideal or practical capacitors. The latter do have some leakage that might include the package, therefore physical limits of the materials already provide the closed circuit.
MrAl:
--- Quote from: wraper on August 05, 2019, 11:48:34 am ---
--- 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 ---
The question does not include anything that will close this (ideal) circuit. Knowing what voltage is there does not require connecting voltmeter. Not to say there are multimeters with with >10 GoHm input resistance with minimal impact on circuit. 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 ---
It is common in theory to call the capacitor voltages *initial*.
That means before anything else happens.
10GOhm, 100GOhm, 10^99 Ohm, doesnt matter, it's still going to cause a change even though small.
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.
This open circuit is not good for anything until it is used for something, and then things change. If 1us later anything is measured, the change starts. If 1ms later, then the change starts then. If 10^99 years later something is measured, then the change starts then. So it's common to call the cap voltage *initial*.
wraper:
--- 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.
MrAl:
--- 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.
wraper:
--- 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.
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