7
\$\begingroup\$

So my friend and I were working on this project which required capacitors. We needed to charge a capacitor with the following rating: 100microFarad, 400V.

When we connected it to a 220V AC source (peak voltage 311V) it went boom. We thought we had somehow shorted the circuit and tried once again making sure that the connections were proper. Once again it went boom. What were we doing wrong? Our setup was simply a few wires and our capacitor.

EDIT

So I got why it blew up (the current surged). Now I'm editing the question to ask this:

So while we were in school, our physics laboratory assistant called up an electrician from downstairs to show how to do stuff. The electrician brought a ceiling fan capacitor and showed us how to connect it to the socket. Maybe he thought we were dealing with such capacitors too. So he charged the capacitor and took it out and touched its two ends in such a way that there was sparking (proof of charging). Why didn't this capacitor blow up? Current must have surged in this one too right? (Thank God it didn't blow up! The guy was holding it in his palms). There wasn't any considerable difference in capacitance (it was 1000 microfarad against the 100microfarad we used) so why didn't it?

\$\endgroup\$

migrated from physics.stackexchange.com Mar 18 '17 at 8:37

This question came from our site for active researchers, academics and students of physics.

  • 10
    \$\begingroup\$ If you connected the capacitor directly to unrectified AC then you'd get a large current though it and that's probably what destroyed it. Actually I'm surprised it didn't blow the circuit beaker in your house. \$\endgroup\$ – John Rennie Mar 18 '17 at 8:34
  • 25
    \$\begingroup\$ @Kunai Pawar It is up to YOU to design in safety systems. The first part of that is to at least have a basic understanding of electricity, which I hazard to guess, you do not. \$\endgroup\$ – R Drast Mar 18 '17 at 10:00
  • 14
    \$\begingroup\$ @KunalPawar Please, please, stop doing anything with high voltage until you've learned enough about electronics to understand why what you did was a really really bad idea. You're going to hurt yourself or someone around you. \$\endgroup\$ – Sneftel Mar 18 '17 at 16:33
  • 2
    \$\begingroup\$ @KunalPawar Your lab assistant may have known what was going to happen and told you to try it to see you jump. Sadly, some people like to explode old capacitors (at a lower voltage!!!) for fun because of the pop they make. \$\endgroup\$ – simpleuser Mar 18 '17 at 21:53
  • 2
    \$\begingroup\$ What did you expect to happen? How much current did you expect to flow? If your answer is that you have no idea, you should not be messing with sources capable of producing significant amounts of power. \$\endgroup\$ – David Schwartz Mar 19 '17 at 7:16
30
\$\begingroup\$

Your 100uF 400V capacitors were undoubtedly low cost aluminum electrolytic capacitors rated only for use on a DC circuit. When you connect such capacitor to the AC mains you are alternately subjecting the component to a positive peak of the 311V and then -311V at a rate of some 50 or 60 times per second. The polarized plates in the electrolytic capacitor very quickly break down under these conditions and can lead to an internal short.

Even if the capacitor plates were able to survive the negative voltage for a short time the effective AC impedance of a 100uF capacitor connected to the AC mains without many other series components to limit the current would result in a goodly amount of power being dissipated in the component resulting in the sealed can expanding and exploding.

\$\endgroup\$
  • 2
    \$\begingroup\$ Its like putting a ~30 ohm resistor across a 220 v line. \$\endgroup\$ – Tejas Kale Mar 18 '17 at 9:42
  • 27
    \$\begingroup\$ I put a 30 ohm resistor across a 230V line. It's part of my stove, it makes great toasted cheese sandwiches. \$\endgroup\$ – Brian Drummond Mar 18 '17 at 11:04
  • 4
    \$\begingroup\$ Yeah, That 30ohm resistor or capacitor in this case is being asked to withstand about 1763 watts of power. \$\endgroup\$ – Michael Karas Mar 18 '17 at 13:55
  • 4
    \$\begingroup\$ If we overlook the capacitor plate degradation for a moment, it's not like putting a 30 ohm resistor across the line, at least not in one sense. I would indeed pass ~7.7 A and the apparent power would be ~1.8 kW but most of that is reactive power and would be lost in the power company's equipment. The real power is P=I²R for the given current, and the ESR of the capacitor. This is still bad news for the capacitor, but a lot less power than ~1.8 kW is emitted through the capacitor. \$\endgroup\$ – nitro2k01 Mar 19 '17 at 5:30
  • \$\begingroup\$ @MichaelKaras I've edited my question. \$\endgroup\$ – Kunal Pawar Mar 20 '17 at 16:19
12
\$\begingroup\$

If it was a polarized capacitor, remember AC will be the wrong polarity half the time, so it will go boom.

If it was a non-polarized capacitor, at 50Hz its impedance is 1/(2πfC) so about 32 ohms, which means 7 amps at 230V, so if it's not rated for this current it will overheat.

\$\endgroup\$
  • 1
    \$\begingroup\$ How violent failure would 7 A cause? P = R_ESR * I^2. At 100 milliohms, "only" about 5 Watts will be converted to heat in the capacitor, causing a relatively slow combustion. \$\endgroup\$ – Oskar Skog Mar 18 '17 at 12:43
  • 4
    \$\begingroup\$ If it's a baddass motor start cap, won't fail. If it's a non-polarized electrolytic... I'd say a slow poof... but my bet is on a polarized electrolytic. \$\endgroup\$ – peufeu Mar 18 '17 at 12:55
5
\$\begingroup\$

Someone stated that putting an AC-rated capacitor across a 220V line would be like putting a 30Ohm resistor across it. That is very much incorrect with regard to the dissipated power but correct with regard to the peak currents. The actual dissipated power is just due to leakage and finite resistance. The bulk of the current flowing in and out of the capacitor is out of phase with the voltage and consequently energy is getting pumped in and out of the capacitor without actually getting dissipated (apart from lossage).

So the power lines in your house (and leading there) are exercised as if you were powering a stove, but your power meter will register rather little net power. This kind of highly capacitive load consequently is quite unpopular with the power companies as they sell rather little energy but still have to pay for all the current flowing through the lines.

At any rate, it does not sound like you know what you are doing, and you are dealing with lethal amounts of energy. Exploding electrolytic capacitors contain corrosive liquids, the fumes are corrosive and unhealthy as well.

Also, in general, if something explodes / catches fire / etc., unless you have a really good idea of what caused it, a good rule of thumb is to try to learn a bit more about what happened before trying a second time.

High voltage experiments without proper experience is a good recipe to kill yourself and/or cause damage to the building and/or its flooring and furniture.

\$\endgroup\$
  • 1
    \$\begingroup\$ Can we please be nice? This answer reads like an insult to the OP. electronics.stackexchange.com/help/be-nice \$\endgroup\$ – bwDraco Mar 19 '17 at 15:53
  • \$\begingroup\$ It read like some reasonably warranted tough love to me. I've suggested an edit that is somewhere between the OP and RoyC's; restoring the original points but still toning it down. \$\endgroup\$ – Jason C Mar 19 '17 at 16:31
  • \$\begingroup\$ @user142646 Indeed I was trying to kill myself. I'll make it a point to succeed the next time:) \$\endgroup\$ – Kunal Pawar Mar 27 '17 at 19:47

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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