0
$\begingroup$

This is a question that's been bugging me for a while. Suppose you have two capacitors which have different capacitances. And they're preloaded with different voltages.

When they are connected in parallel, how do the electrons flow so that they end up having the same potential drop?

Sorry for my not-so great English and thanks in advance!

Improve this question
$\endgroup$
5
  • $\begingroup$ Can you be more specific on what is bugging you and how is it different from the case with 2 resistors in parallel? $\endgroup$
    – Ofek Gillon
    Commented Feb 11, 2017 at 9:09
  • $\begingroup$ Which situation are you asking about: (1) two capacitors in parallel start uncharged and the voltage is increased, (2) two capacitors in parallel are at a constant voltage, or (3) two capacitors start with different voltages and are then connected to each other? $\endgroup$
    – LedHead
    Commented Feb 11, 2017 at 9:38
  • $\begingroup$ Two capacitors are charged separately and then connected in parallel. $\endgroup$
    – finnishStudent
    Commented Feb 11, 2017 at 9:41
  • $\begingroup$ @finnishStudent Could you edit the question to reflect that? I think you say "capacitances" when you mean "voltages". $\endgroup$
    – LedHead
    Commented Feb 11, 2017 at 9:45
  • $\begingroup$ No I mean C aka capacitance :) $\endgroup$
    – finnishStudent
    Commented Feb 11, 2017 at 9:47

2 Answers 2

Reset to default
2
$\begingroup$

If you have two charged capacitors they will each have a potential difference $V_1$ and $V_2$ across their plates.

If one plate of one capacitor is connected to one terminal of the other capacitor with a conductor those two terminals will be at the same potential.

This means that the terminals which are not connected have a potential difference of $V_1-V_2$ between them.
If those two terminals are connected together, because you now have a situation where the potential difference across one capacitor is not the same as the potential difference across the other capacitor, current will flow between those two terminals (with an equal current between the other two terminal) until the potential difference across both set of capacitor terminals is the same.

Update in answer to a comment

Here is a numerical example to illustrate the points that I have made and also to show the movement of the electrons during the charge rearrangement process.

enter image description here

Improve this answer
$\endgroup$
6
  • $\begingroup$ Okay, how do the electric fields inside capacitors change when they're connected in parallel? $\endgroup$
    – finnishStudent
    Commented Feb 11, 2017 at 9:18
  • $\begingroup$ The electric fields between the plates of the capacitors depend on the charge on each plate as well as other factors. As the charges move between the two capacitors the electric fields between the plates of the capacitors change until the potential difference across each of the capacitors is the same. $\endgroup$
    – Farcher
    Commented Feb 11, 2017 at 9:27
  • $\begingroup$ Okay in which direction do they flow. Because my physics textbook says electrons flow from the capacitor that has lower potential to the one that has higher potential and this somehow lowers the potential of the capacitor that has higher potential. But this is confusing because doesn't the addition of electrons increase the potential between the plates? Sorry for this long comment. $\endgroup$
    – finnishStudent
    Commented Feb 11, 2017 at 9:33
  • $\begingroup$ @finnishStudent I have updated my answer. $\endgroup$
    – Farcher
    Commented Feb 11, 2017 at 9:56
  • $\begingroup$ An important aside here is that this can't really happen in real life with ideal conductors and capacitors. If you connected two ideal capacitors with different voltages via ideal wires, it would result in an instantaneous change in voltage which requires infinite current, since $I=C\frac{dV}{dt}$. In reality, something's gotta give. Maybe the voltage doesn't instantaneously change, maybe one of the capacitors goes "pop". A similar question $\endgroup$
    – LedHead
    Commented Feb 11, 2017 at 10:10
0
$\begingroup$

You have a load $Q_1=U_1 C_1$ on the first capacitor, and a load $Q_2=U_2 C_2$ on the second capacitor. The combined capacitors have a capacity $C_1+C_2$, so you'll end up with a voltage of $V=\frac{U_1C_1+U_2C_2}{C_1C_2}$. So far, so nice, but the total stored energy is different. So at some point in that summary view of charges and capacity, we did a bit too much of handwaving.

Now if we connect those capacitors and consider the charges/voltages to immediately adjust, infinite current has to flow. "Infinite" is not an option even when dealing with very good components, so we have to take into account that the connection between the capacitors will have a bit of resistance and a bit of inductivity and is immersed in the ether (well, there is no such thing but empty space is good enough for EM waves). As a result, we have a dampened oscillator which will come to rest (with the voltage we have calculated) when the missing energy has been dissipated. Assuming the connection is solid enough, most of it will be very quickly radiated as RF rather than heat and if you have a radio running nearby, it will likely produce a click at that moment.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.