Charging Capacitors: How Does Series Ckt Work?

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Discussion Overview

The discussion revolves around the behavior of capacitors connected in series to a power source, specifically addressing how they charge over time, the implications of differing capacitance values, and the resulting charge distribution among the capacitors.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant notes that when a capacitor is connected to a power source, it eventually becomes fully charged, leading to no current flow. They question how this behavior changes when multiple capacitors with different capacitances are connected in series.
  • Another participant explains that capacitors in series behave as a single capacitor with a combined capacitance calculated using the formula 1/Cnew = 1/C1 + 1/C2, resulting in a total capacitance that is always less than or equal to the smallest capacitor in the series.
  • It is asserted that all capacitors in series will accumulate the same charge, but the potential difference across each capacitor will vary inversely with their capacitance. This means that capacitors with higher capacitance will have a lower potential difference.
  • Concerns are raised that if current continues to flow, capacitors with lower capacitance may fail before those with higher capacitance can reach their full charge, potentially leading to uneven charging among the capacitors.

Areas of Agreement / Disagreement

Participants express a shared understanding of the basic principles of capacitors in series, but there are differing views on the implications of varying capacitance and the effects of continued current flow on charging and potential failure.

Contextual Notes

Some assumptions regarding the behavior of capacitors under continuous current flow and the conditions for failure are not fully explored, leaving room for further discussion on these points.

xoxo1001
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From my understanding, if I connect a capacitor to a power source, given enough time the capacitor will be fully charged and no current goes through the circuit. What if I have two or multiple capacitors connected in series? Do they all get charged at the same time? What if they have different capacitance? Does the capacitor with the lowest capacitance get charged first? And if it does get charged, there shouldn't be any current going through, so does that mean the other capacitors with higher capacitance won't get full charge? I'm a bit lost.
 
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What happens is that the capacitors essentially form one big capacitor with the following capacitance:

1/Cnew = 1/C1 + 1/C2

Meaning, the resulting capacitance is always smaller or equal than the smallest capacitor in the series.
 
Because they are in series, the capacitors will each accumulate the same charge. Because they have different capacitance, this means that the potential difference across each will be different -- in inverse proportion to their capacitance.

A "fully charged" capacitor is one that has as much charge (and therefore as much potential difference) as it can hold without failing. So yes, this will mean that the capacitors with the higher capacitance will have a lower potential difference and will likely not yet be near their failure point and so will not be "fully charged".

If you keep pumping current through the circuit you will cause the capacitors with lower capacitance to fail before the capacitors with a greater capacitance can reach their full charge.
 
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jbriggs444 said:
Because they are in series, the capacitors will each accumulate the same charge. Because they have different capacitance, this means that the potential difference across each will be different -- in inverse proportion to their capacitance.

A "fully charged" capacitor is one that has as much charge (and therefore as much potential difference) as it can hold without failing. So yes, this will mean that the capacitors with the higher capacitance will have a lower potential difference and will likely not yet be near their failure point and so will not be "fully charged".

If you keep pumping current through the circuit you will cause the capacitors with lower capacitance to fail before the capacitors with a greater capacitance can reach their full charge.
Ah ok! That makes sense! Something clicked! Thanks very much!
 

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