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Capacitors charged in parallel, discharged in series - effect on current?

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Forgive the basic question, but if I were to charge up capacitors in parallel and then reconnect them in series, and discharge them, should the current remain the same (as the charging current) and the voltage multiply by the number of capacitors?
 

If the charging current is :

  • ic = C . dV/dt
The discharge current of the capacitors rearranged:

  • idc = (1/n)C . d(nV)/dt

Seems like being the same current.
 
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andre_teprom said:
If the charging current is :

  • ic = C . dV/dt
The discharge current of the capacitors rearranged:

  • idc = (1/n)C . d(nV)/dt

Seems like being the same current.
Click to expand...

Consider a concrete example. You take two capacitors of equal value. You charge them to a potential of V. Without any series resistance, both with instantly charge with an infinite current to the final potential V. Let us put for fun some series resistance (equal for both). As the resistances are in parallel, the initial current will be V*2/R.

Remove the potential souece but do not touch with hand the free ends and put the capacitors in series. The voltage will be 2V and the series resistance will be 2R. The initial discharge current will be V/R.

Not quite the same thing.
 

Viewed in this perspective, you're right. I was considering that the original concern was to the change on current flowing accross each capacitor, but the total current supplied by power source is really different, being n times greater in the charging.
 
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The discharge current depends upon the resistance of the load connected across the two caps in series.
It has nothing to do with the charging current.
 

The question as posted in fact lacks of further details, mainly whether it refers to some specific application with switched capacitors in power electronics, or if it is just an hypothetical scenario, but seemed like OP was referring to the peak current that the component must be rated in such arrange ( and the maximum value of id or idc would rate the capacitor ).

Anyway, for real world agree with you that there is no practical use in calculating the same current of charge/discharge, in except of a short-circuit at output.
 

Ideally as you have said ,if we take 2 same valued capacitors and charge it upto say 10V and then hypothetically connect them in series then total voltage across will no doubt be 20 V or 2*V but the discharge current depends solely upon the load which will in this case be 2*V/RL wherein RL=Load resistance.
 

crutschow said:
The discharge current depends upon the resistance of the load connected across the two caps in series.
It has nothing to do with the charging current.
Click to expand...
This is the correct answer.

Charging in parallel and discharging in series is a very common way to generate extremely large high voltage dc discharges.
Check out "Marx generator":
https://en.wikipedia.org/wiki/Marx_generator

The Russians built a real beauty, six MILLION volts !
https://www.rt.com/news/181748-tesla-marx-generator-lightning/
 

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