0
\$\begingroup\$

Assume I have two identical capacitors and charged both equally. If I measure C1 with a voltmeter, I read 5V, and C2 reads 5V. If I connect the positive leg of C1 to the negative leg of C2 (connect in series) what would the voltage be across the series capacitors?

Googling series capacitor seems to suggest they add up to 10 volts but I am highly doubtful and somewhat scared to try.

Wouldn't all the holes of the positive C1 leg rush to merge with electrons in the negative C2 leg and quickly deplete the charge, possibility resulting in an explosion if the capacitors are big enough?

Improve this question
\$\endgroup\$
3
  • \$\begingroup\$ so yes 10V until they decay. e-caps are like really weak coin cells and would have to be 1000x bigger roughly than a coin cell to last 24 hr like a coin cell on an LED. The difference is CAPs have low ESR and long life for recharging and of course coin cells are primary chemical voltage sources \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 19 '20 at 6:08
  • \$\begingroup\$ Google "switched capacitor convertor" for a common application of this concept. \$\endgroup\$ – The Photon Dec 19 '20 at 7:05
  • \$\begingroup\$ yea thanks good to know. \$\endgroup\$ – au kk Dec 19 '20 at 8:55
1
\$\begingroup\$

The electrons will not rush to equalize the charge at the connection between the two capacitors, because the charges are held in place by the counter charge on their respective opposing plates.

Similarly there is no explosion or short when you connect multiple batteries in series in order to multiply the voltage, eg. obtain 6V from 4 AA 1.5V batteries.

There are concerns, however, when connecting capacitors in parallel when their voltages are not equal. In that case an equalizing current will flow, and its magnitude is only limited by the (small) resistance of the interconnecting conductor (wire).

There are grave safety concerns when electrolytic capacitors are connected with reverse polarity, so when attempting any of the above, polarity matters even with small voltages.

As an interesting aside, a "voltage multiplier" works on the principle of charging capacitors and then connecting them in series in order to obtain a higher voltage than the supply.

https://en.wikipedia.org/wiki/Voltage_multiplier

Improve this answer
\$\endgroup\$
7
  • \$\begingroup\$ " ... because the charges are held in place by the counter charge ..." yea ok i didnt think about that but there would still be some current as the other cap is linked with a conductor and opposite plate is not. but i guess its not enof to make voltage multiplier circuits useless, correct ?? \$\endgroup\$ – au kk Dec 19 '20 at 8:51
  • \$\begingroup\$ also the battery analogy really doesnt apply. battery does not cotain free charge internally that can move as soon as there is path. the rate of release of the electrons is constrained by how fast the chemical reaction can occur. battery has variable internal resistance depending on current, and max current for batteries is thousands of times less than capacitors if not more. \$\endgroup\$ – au kk Dec 19 '20 at 8:53
  • \$\begingroup\$ The battery: the question was whether a charge would move, and it won't, regardless speed as it is a similar open circuit situation. As for "there would still be some current": the net charge on the capacitor (both plates together) is zero so there will be no current. This is different from ESD between to unequally charged objects. \$\endgroup\$ – P2000 Dec 19 '20 at 18:17
  • \$\begingroup\$ I just can not believe this. I would never believe the no current assertion unless experiment proves me wrong. This is entirely different from battery. It has no such thing as "open circuit similarity". In an open circuit battery the underlying chemical reaction simply can not be completed if the leaving electrons dont return on the other end. the capacitors have free electrons and holes. like a water tank with two holes at the bottom. that the water wants to flow thru the first hole does nothing to prevent it from flowing to the 2nd hole. \$\endgroup\$ – au kk Dec 22 '20 at 19:42
  • 1
    \$\begingroup\$ I'll try to do the experiment and post here, but thanks for the engagement \$\endgroup\$ – au kk Dec 23 '20 at 17:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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