Initially, charge is shared between Source C1 and the two Co capacitors which make up Ctotal. Once charge is stabilized, the wire link is cut, forming a tiny capacitor Cm in series which is added to C total, reducing Ctotal to zero. What happens to the charges and voltages in the loop then, and also when the gap Cm is opened and closed repeatedly?

  • \$\begingroup\$ What makes anyone think that snipping the wire in step 3 causes anything at all to happen? Which is why nothing still happens when it's reconnected in 4. \$\endgroup\$
    – Neil_UK
    Commented Jan 6, 2018 at 8:02
  • \$\begingroup\$ Do you think that the location of the link has any relevance? What’s the difference between opening the link and opening the switch? Remember that there is zero flow of current through the dielectric. \$\endgroup\$
    – Chu
    Commented Jan 6, 2018 at 11:34
  • \$\begingroup\$ +Neil_UK: Snipping the wire effectively adds a tiny capacitor Cm to the compound series capacitor [C0+Cm+Co] reducing C(total) to zero. As +Vinzent said in his answer below, reducing a charged capacitor will increase the voltage across it. So opening and closing the link should make the voltage across C(total) vary from 1/2VC1 to very large, causing charge across C(total) to slosh back and forth between C1 and C(total). If nothing happens, I really want to understand where my logic is breaking down. \$\endgroup\$ Commented Jan 7, 2018 at 0:37

2 Answers 2


Once the capacitors are all charged and no current is flowing in them the cicuit you get is this one:


simulate this circuit – Schematic created using CircuitLab

disconnect the wire between the two capacitors Co and the circuit you get looks like this


simulate this circuit

Since VCo=VC1/2 we now have a voltage between the terminals (split wire ends) of -VCo+VC1-VCo=0v

Now you can of course reconnect the two wire ends, but as there is no voltage difference between them no current will flow.

As Neil sayed a lot of nothing happens.


If you were to actually change the capacitance of a capacitor while it is charged what would happen is that the voltage would change. considder this schenario;

You have a capacitor (two plates opposite each other) charged to some voltage. now you move them appart by some distance, the capacity is going to decrease but because there is the same amount of energy stored in the capacitor the voltage will increase to compensate so that Q=(V^2*C)/2 stayes unchanged.

  • \$\begingroup\$ +Vinzent: doesn't adding the tiny capacitor Cm into the series Ctotal = [Co+Cm+Co] actually change the total capacitance to zero, causing some sort of voltage spiking across C(total)? If not, why not? Thanks \$\endgroup\$ Commented Jan 7, 2018 at 0:47
  • \$\begingroup\$ No nothing happens at all the energy is not stored in the wire so you are not changing anything by Breaking the circuit. You can put all the capacitors into the circuit in series that you want the original caps are still going to have to same charge on Them. If you were to physically change the capacitance of one of the caps the. You could change the voltage, only then \$\endgroup\$
    – user173292
    Commented Jan 7, 2018 at 1:08

To get from diagram 1 to diagram 2, we close the switch. Current flows to redistribute the charge. There's no resistor shown in the diagrams, but practical wiring will have a finite resistance. This will limit the current that flows when the switch closes, and dissipate the excess energy that was contained in C1, when the charge redistributes to all the capacitors.

After a while in diagram 2, the current will have stopped flowing. The wire between the two C0 capacitors will have zero current flowing through it, and the voltage at both ends will be equal.

To get to diagram 3, we snip the wire. As there was zero voltage across its ends, it now has zero voltage across the cut. There was no current flowing in it before the cut. Certainly no current will flow in it after it has been cut. With no current flowing, there is no charge redistribution. Nothing else happens. All capacitors stay charged to the voltage they were at in diagram 2. All capacitors maintain the same charge as they had in diagram 2.

To get to diagram 4, we reconnect the cut. There was no voltage across it before reconnection, nothing will happen when it's reconnected. Nothing else happens.

If we cut and reconnect the loop many times, then nothing will happen, many times.

  • \$\begingroup\$ Here's an interesting question: would it take extra work to open and close the link, apart from the trivial amount of mechanical force need to actuate a switching mechanism? Do the electromagnetic forces somehow make it more difficult? Someone here (aka: "xox") seems to think so: forum.allaboutcircuits.com/threads/… Thread was closed before I could ask for clarifications. Seems they didn't want to discuss it any further. \$\endgroup\$ Commented Jan 7, 2018 at 0:52
  • \$\begingroup\$ No, it doesn't take any extra work. \$\endgroup\$
    – Neil_UK
    Commented Jan 7, 2018 at 7:36

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