Two capacitor problem (open circuit version) - Page 1
Forum section: electronics / beginners

author: Meshka7
author_rank: Contributor
timestamp: July 27, 2019, 07:13:03 am
content: What happens when you charge a capacitor to a 1000V then connect another discharged capacitor to the 1000V terminal (without completing the circuit)? What will be the potential on both sides of the discharged capacitor? will it get charged?

author: Yansi
author_rank: Super Contributor
timestamp: July 27, 2019, 07:20:05 am
content: No, why should it? The circuit is incomplete.

author: Shock
author_rank: Super Contributor
timestamp: July 27, 2019, 12:50:22 pm
content: Best to draw out the schematic in both states and mark where your measurement is across. Also you can test this with two low voltage capacitors or a circuit simulator it's a lot safer, unless you are talking about dielectric absorption or something else to do with a high voltage capacitor.

author: Meshka7
author_rank: Contributor
timestamp: July 27, 2019, 06:51:58 pm
content: No, it's a theoretical question. Not planning to try it out.floating A------|1000V|----------|Discharged|----------floating BThe question is: What is the voltage at point B?

author: wraper
author_rank: Unknown
timestamp: July 27, 2019, 06:54:59 pm
content: Quote from: Meshka7 on July 27, 2019, 06:51:58 pmpoint A (reference point)----------|1000V|----------|Discharged capacitor|----------point B (1000V)1000V, since voltage across discharged cap is 0V.

author: RoGeorge
author_rank: Super Contributor
timestamp: July 27, 2019, 07:52:09 pm
content: The second capacitor will charge at 500V if the two capacitors are the same.If the two capacitors are not the same, you will have a capacitive voltage divider with V2=V1*C1/(C1+C2).

author: wraper
author_rank: Unknown
timestamp: July 27, 2019, 09:01:01 pm
content: Quote from: RoGeorge on July 27, 2019, 07:52:09 pmThey are not connected in parallel. Circuit is open, so charged capacitor will remain charged and discharged capacitor will remain discharged.

author: Meshka7
author_rank: Contributor
timestamp: August 03, 2019, 12:40:45 am
content: So if you replace the 0V capacitor with a Mosfet, will it get damaged? (given that it has 0V across it?)

author: MrAl
author_rank: Super Contributor
timestamp: August 05, 2019, 05:58:30 am
content: 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.

author: wraper
author_rank: Unknown
timestamp: August 05, 2019, 11:48:34 am
content: Quote from: MrAl on August 05, 2019, 05:58:30 amThe 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.

author: SparkyFX
author_rank: Frequent Contributor
timestamp: August 05, 2019, 12:09:32 pm
content: 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.

author: MrAl
author_rank: Super Contributor
timestamp: August 05, 2019, 01:31:49 pm
content: Quote from: wraper on August 05, 2019, 11:48:34 amIt 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*.

author: wraper
author_rank: Unknown
timestamp: August 05, 2019, 02:34:58 pm
content: Quote from: MrAl on August 05, 2019, 01:31:49 pmThis is not real circuit but theoretical question.QuoteThese 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.

author: MrAl
author_rank: Super Contributor
timestamp: August 06, 2019, 08:54:14 am
content: Quote from: wraper on August 05, 2019, 02:34:58 pmUseless to you not to me.

author: wraper
author_rank: Unknown
timestamp: August 06, 2019, 12:54:29 pm
content: Quote from: MrAl on August 06, 2019, 08:54:14 amI mean don't add things which do not exist in the question. 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.

author: MrAl
author_rank: Super Contributor
timestamp: August 07, 2019, 03:36:40 am
content: Quote from: wraper on August 06, 2019, 12:54:29 pmIt'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.

author: wraper
author_rank: Unknown
timestamp: August 07, 2019, 10:31:40 am
content: Quote from: MrAl on August 07, 2019, 03:36:40 amLook 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 amis pointless becauseQuote from: wraper on August 05, 2019, 11:48:34 amWith any circuit with these capacitors initial voltage will be the same. But what will happen later will be very different.