# Putting two ends of a capacitor together

Discussion in 'General Electronics Chat' started by pprendeville, Jan 13, 2010.

1. ### pprendeville Thread Starter New Member

Nov 19, 2009
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What would happen if I were to put the two wires of a capactor together when it's charged? The capacitor in question is rated for 50V and 220microF.

Also, what are the effects of setting a multimeter to ohms and hooking up probes to the wires of a capacitor (+ve to +ve, -ve to 0ve and vica versa).

I've measured the voltage of the capacitor and it seems to be a 0.45V charge in it.

2. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
The capacitor would immediately discharge if you shorted its' terminals together. This is common practice when working on equipment that may have high voltage stored in capacitors. A test lead with a probe is connected to a solid chassis ground, and the capacitors are probed to discharged them, making the circuit safe to work on.

Don't attempt to connect a meter set to Ohms OR mA's to a capacitor unless you are certain that the capacitor is discharged, or you will very likely damage the meter.

Measure the voltage across the cap, and unless it measures 0v, discharge it by shorting the terminals. Then you can experiment with measuring the resistance.

The meter outputs a voltage when measuring Ohms. This will begin to charge the capacitor. Old-style d'Arsonval movement meters (the ones with the needles) are handy to see if capacitors are internally open or shorted; you can see the cap charging by the needle moving from 0 Ohms to nearly infinite.

3. ### pprendeville Thread Starter New Member

Nov 19, 2009
15
0
Me being a novice, I'm not very familiar with some of your terminology so is it ok if I ask some questions that may appear dumb.

Would a resistor be used more commonly to discharge capacitors? What would high voltage be? By this I mean would 50V be considered high voltage or would it need to exceed a certain amount before its considered high voltage? Is it the voltage or the current in a circuit that can potentially be fatal or both?

Ive measured the capacitor voltage and it now reads 0.29V. So its almost fully discharged right? Is it safe now to use it with the ohmmeter? Also, how would I know if I damaged a meter? Would I get faulty readings like open circuit when its obviously closed etc?

What would happen if the capacitor was discharged and the terminals were shorted?

Is this particular capacitor a danger to me if I was using it incorrectly or would there be just a small pop if it blew? Thats enough silly questions for one post I think.

4. ### Darren Holdstock Active Member

Feb 10, 2009
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Yes, discharge with a resistor or the cap may be damaged by the current surge (some types of cap are more easily damaged than others). If you dead-short it then almost all the energy stored in the cap will be suddenly dumped in the equivalent series resistance (ESR), which can cause localised burnouts and fractures. That energy, in joules, is (CV^2)/2.

Here's a fun introduction to the phenomenon of dielectric soakage: Charge and then discharge a cap (cheap electrolytics are good), immediately connect a voltmeter across it, and watch the voltage creep back up a little. In the olden days when TVs had CRTs, repair engineers would have to short out the anode terminal at least 2 or 3 times before it was safe to touch.

Jul 7, 2009
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When I was a student, one of the teachers used to have me come into his class and demo the discharge of a reasonably large bank of capacitors. I'd charge it up with a big old vacuum tube power supply to a few hundred volts. Then when I got the signal from the teacher, I'd short across the bus bars with my Xcelite screw driver. The noise was rather surprising to everyone (it was quite loud) and made a good demo of the energy that could be stored (the teacher would talk for a fair while after the capacitor bank was charged). I still have that orange-handled screwdriver somewhere around here with a bunch of dings taken out of its shank.

6. ### beenthere Retired Moderator

Apr 20, 2004
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And those of us who wanted no surprises left a small screwdriver stuck under the connector with a clip lead to the chassis.

7. ### peranders Well-Known Member

May 21, 2007
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The phenomena is called dielectric absorption which means that the charge is a little bit "sticky" and won't leave the cap.
http://www.national.com/rap/Application/0,1570,28,00.html