If I fully charge a 1.5V capacitor from 1.5V battery, how long can it keep the charge? If I connect it to 1.5V bulb would it lose that charge instantly? If yes why? Where would it go?

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    \$\begingroup\$ What is a 1.5 bulb? \$\endgroup\$ – Kellenjb Jan 25 '12 at 15:09
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    \$\begingroup\$ Just an observation: I see that you are asking many basic questions; maybe you could search for a book to get into concepts... \$\endgroup\$ – clabacchio Jan 25 '12 at 15:57
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    \$\begingroup\$ This topic is well described in Art of Electronics book. \$\endgroup\$ – Al Kepp Jan 25 '12 at 16:07
  • \$\begingroup\$ Or electronics.stackexchange.com/questions/616/… \$\endgroup\$ – clabacchio Jan 26 '12 at 10:09

In case of discharge through a resistance, there is an exponential law: $$ V = V_{0} \cdot {e^{-t \over RC}} $$ but to get the time you need R and C: you have provided only V.

This figure shows it in a graphical way (for V=1V and RC~=1):

Discharge of a capacitor (courtesy of cy.wikipedia.org)

In your case, it will lose the charge creating a current over the lightbulb which will consume power (equal to VxI) to produce light for your eyes :) (and heat). The lower the bulb resistance, the greater the current supplied, and the shorter discharge time.

Note that the current will have the same curve of the voltage, as their ratio is given by the resistance.

This case is appliable also without load, if you know the equivalent resistance between the two plates (equivalent leakage resistance).

And this is only the basic transient, with R fixed. If you have a constant current, so it becomes a linear discharge with law: $$ V={Q \over C} = {Q_0 - t \cdot I \over C} $$


If you connect a charged capacitor to a light bulb or other device that consumes current, the charge will flow through the device. It won't happen "instantly", but depending upon the size of the capacitor and the amount of current drawn by the device, it may happen in less than a millisecond. Note that because most devices will draw less current as the voltage falls, the rate at which the capacitor discharges will fall as well. When connected to a purely resistive device, the capacitor voltage will fall by a factor of e (about 2.71828) every RC seconds, where R is measured in ohms and C in farads (or megs and microfarads).

If there is no load on a cap, in the absence of chemical breakdowns or other such factors, it will only lose current through internal leakage. While some caps have sufficient leakage that they will lose much of their charge in a matter of hours, a good quality cap may be able to maintain 90% or more of its charge for a period of years. The biggest difficulty keeping a cap charged that long is avoiding leakage; 4.5 megs of leakage resistance would drain a one-farad cap about halfway in a year.


A capacitor can keep its charge indefinitely (in theory). That's why with large capacitors it is dangerous to open high voltage equipment even years after they have been disconnected. What you are probably asking is the time the capacitor needs to discharge. It will discarge according to an exponential law. In the case you described (a capacitor feeding a bulb), you will have an RC circuit.

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    \$\begingroup\$ every capacitor has a leakage current and it is often larger then people realize on large capacitors, making them not that dangerous in the long long term. \$\endgroup\$ – Kortuk Jan 25 '12 at 18:32
  • \$\begingroup\$ @Kortuk: that's right, that's what I meant by 'in theory'. :) \$\endgroup\$ – Count Zero Jan 25 '12 at 20:18
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    \$\begingroup\$ Noting very old equipment is still dangerous is in practice. I felt it was worth noting in that case. \$\endgroup\$ – Kortuk Jan 25 '12 at 20:52

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