Resistance of a discharging capacitor

  • Thread starter Pharrahnox
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  • #1
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I am trying to calculate the maximum current drawn from a capacitor that is connected to a resistor of a very low value (a few mΩ). Because the resistor does not have a large resistance as I've seen with many V/t and I/t graphs for capacitors connected to resistors, where the decay of the curve depends on the RC time constant, I am pretty sure that the resistance of the capacitor would have a significant affect on the maximum current.

How can this resistance be determined? If the resistance of the capacitor changes over time, which I'm pretty sure it does, how can I find its lowest resistance from which current is drawn?
 

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  • #2
Drakkith
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I don't think the resistance changes over time. In order to get an accurate decay curve and find the current you would need to find the total resistance of the circuit, which will include the resistance of the resistor, the capacitor, and the wires. The latter two are usually so small that we can ignore them, but if your resistor has very little resistance then they may become significant factors.
 
  • #3
Philip Wood
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Another complication, if the resistance of the capacitor is significant, is that it won't behave like a single resistor in series with a perfect capacitor, because the resistance will be distributed over the area of the capacitor plates. A better model might be a ladder filter with resistors as the 'horizontal' elements and capacitors as the 'vertical' elements.

But perhaps I'm scaremongering. No doubt the circuit could be investigated with some sort of data-logger that samples voltages at very frequent intervals, and can therefore deal with very rapid charge or discharge.
 
  • #4
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Do you know the make and model of the capacitor?

Manufactures often publish extensive amounts of data for electronic components. You can usual find an effective serial resistance.

The other option is to build a simple RC circuit and measure the decay time.
 
  • #5
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According to this site: http://www.chemi-con.com/components/com_lcatalog/uploaded/8/3/3/48990562506224c250b19.pdf [Broken] at pg26 63V 12000μF capacitors have about 21mΩ ESR.

I have 2 of these, so I could connect them in parallel to get a total capacitance of 24mF. Since they are in parallel, does this then mean that the ESR would be halved (since there's two of them)? If so, this would mean that the maximum current would be assuming the wire (in this case acting as a resistor) resistance to be 1.4Ω, then the total resistance would be 1.41Ω, and the maximum current at 60V would be 42.55A.

However, there would be other factors that would drastically affect that value, wouldn't there? And if so, roughly how much of an affect would they be expected to have on the maximum current?
 
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