What happen if a charged capacitor is removed from a battery? In other words, first, by using a battery, charge a capacitor. And then, remove the capacitor from the circuit. If we do this, what happens? Is it dangerous? Please explain the result specifically!

  • 4
    \$\begingroup\$ Please provide a more detailed question, including a schematic or diagram of the test setup and what action is performed during the test. I think you mean that the cap is in parallel with the battery, then it is disconnected from the battery. If so, please edit the question to reflect that. \$\endgroup\$
    – user57037
    Dec 27, 2017 at 5:17
  • \$\begingroup\$ I added some details in my question according to your opinion:) \$\endgroup\$
    – user172943
    Dec 27, 2017 at 5:35
  • \$\begingroup\$ If the capacitor and battery are connected long enough to have the same voltage, then removing the capacitor should not be dangerous. What I mean is, there would not be a spark or anything. Of course, all batteries and all charged capacitors are potentially dangerous, depending on the voltage and how much energy is stored. But I don't think that is what you are asking about. \$\endgroup\$
    – user57037
    Dec 27, 2017 at 5:45
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    \$\begingroup\$ Connecting a discharged capacitor to a battery can cause very large surge currents. And if the connection is made by a switch with bounce, then there will probably be arcing at the switch due to the high initial current. But opening the circuit after the capacitor is charged up should not cause any spark. \$\endgroup\$
    – user57037
    Dec 27, 2017 at 5:47

1 Answer 1


The capacitor will start discharging because of leakage current. Leakage current is the unwanted current that that exist between the two plates. For general purpose we ignore this leakage current as it does not come into play. But if you are considering long delays, or storing charge in capacitor then it will come into play.

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You can read more in this article which says

The best capacitor I tested lost 5% of its voltage overnight; the worst lost 50% in 5 minutes.

Further Explanation

We know that Q = CV => V = Q/C

Also I=Q/t =>Q=It

V= (It)/C

Note that (It) is the accumulated charge (Q) on the capacitor (Q = It). Now with passage of time this Q will decrease because of leakage current, but the capacitance of the capacitor will remain the same, hence decrease in voltage over passage of time.

Is it dangerous

Yes. you should never remove a charge capacitor from a circuit but voltage below 12V are generally safe. At low voltage human body resistance is high enough so there is no danger of electrocution but as voltage increases, human body resistance decrease which is the reason why high voltage tasers are used for electric shock.

How dangerous they are depends on the voltage and the charge the capacitor holds. High voltage help reduce resistance of human body and the charge (Q) maintains the current in the body. 1 mA current usually produces sensation of electric shock in the body. I won't go into detail but when the voltage is high enough so it can drive 1mA+ current in human body, you will feel sensation of electric shock.

  • \$\begingroup\$ What is the accurate meaning of the final sentence? Could you explain more specifically? :) \$\endgroup\$
    – user172943
    Dec 27, 2017 at 5:38
  • \$\begingroup\$ I have added explanation. Because of decrease in charge Q over time, voltage decreases. \$\endgroup\$
    – TheTechGuy
    Dec 27, 2017 at 5:57
  • \$\begingroup\$ Good complete answer plus one from me, let’s hope the OP recognises the quality of the answer. \$\endgroup\$
    – Solar Mike
    Dec 27, 2017 at 16:25
  • \$\begingroup\$ And charged capacitors , such as used in car distributors are great when thrown across the workshop when shouting “catch” - of course we never did that when we were apprentices... :) \$\endgroup\$
    – Solar Mike
    Dec 27, 2017 at 16:26

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