# Redistribution of charges when capacitors are connected in series

This question is asked before but I didn't get the answers

It is known that when 2 unequally charged capacitors are connected in parallel then the charges redistribute themselves till the voltage across each capacitor becomes equal.

Now if I take 2 capacitors connected in series of capacitance and voltage across each of them (C1,V1) and (C2,V2) respectively such that V1>V2 then what will happen? Will the charges move from higher to lower potential till the potential gets equalised like in case of redistribution of charges in capacitor in parallel combination OR there will be something else?

If something else, then please suggest a appropriate answer. Thanks

• Your second paragraph is only true for capacitors in parallel - but your question is about series. Please clarify your question. What are the initial conditions of the two capacitors before they are connected in series? For your third paragraph - are those (V1, V2) the initial conditions? Why do you think they should end up with equal voltage? The confusing way you phrase the question is probably a result of your own confusion. If you can make the question VERY clear, the answer will become obvious (or at least the question becomes answerable). – Floris Sep 12 '16 at 16:52
• This question is not well posed. It contains false statements. – freecharly Sep 12 '16 at 18:07
• Sorry for the not posting a good question. Hope now the post is clear to all of you. – Perspicacious Sep 12 '16 at 18:23

## 2 Answers

Although your Question is Not Clear but I am assuming that you want to know if there is any charge redistribution in Connecting Capacitors in Series or not. There is no such redistribution of charge in series connection of capacitors. The Charge remains same but something has to rearrange and hence the redistribution of Potential Difference takes place. As There is inverse relation, hence Potential difference redistributes in inverse ratio of capacitance. Hope this helps

For capacitors in series: the same current flows through both of them for the same time. So the electric charge Q is the same on each of the capacitors. Use the capacitor formula Q = CV to calculate the potential on each capacitor in terms of the unknown but equal Q's. These 2 voltages add up to the applied voltage.

For example if one capacitor is 4x the size of the other; then you will discover that the smaller capacitor has 4x the voltage as the larger capacitor. If the applied potential is 250v then the large capacitor has 50V and the small has 200V.

In very high voltage circuits one is sometimes forced to use a set of equal capacitors in series as a smoothing circuit; to remove hum. It is imperative to connect high resistance resistors (say 100 M ohms) across each of the capacitors, to equalize the leakage currents and ensure that the capacitors share the high potential equally.

To answer your question: Connect the capacitors in series and place your load resistor across the two ends. Current will flow until one of the capacitors is empty with zero volts across it. But the voltage on the remaining capacitor will keep current flowing so the empty capacitor receives a reverse charge; thus developing a negative voltage. When the positive and negative voltage become equal; the current ceases to flow and the charges can be calculated using the starting conditions and Q = CV keeping in mind that the same current has flowed through both for the same time giving the same change in Q. Set up the simultaneous equations and solve them to get your voltage and your charges. PS: always draw a Before and After diagram with the appropriate values - known and unknown - so that you can easily see what the equations should be.