Charging Capacitors: How Does Series Ckt Work?

[xoxo1001]

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.

 

[rumborak]

What happens is that the capacitors essentially form one big capacitor with the following capacitance:

1/C_new = 1/C₁ + 1/C₂

Meaning, the resulting capacitance is always smaller or equal than the smallest capacitor in the series.

 

[jbriggs444]

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.

 

[xoxo1001]

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!