Capacitor discharge speed

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Jo-W

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I have a capacitor of 30.000 uF, 35VDC

I am using a voltage of 30V, the capacitor is connected to a coil which has a resistance of 6 ohm.

This means my capacitor is fully charged at: 30V * 0,03F = 0,9 Coulomb.
With 6 ohm, my coil will use 30V / 6 hm: = 5 A.

And 0,9C / 5A = 0,180 seconds.
My question is: does this mean my capacitor will discharge in 0,18 seconds? Are there any other factors involved? capacitor resistance? -log() discharge speed? etc


Thanks
 
It will also depend on the inductance of the coil.
 
Well for one thing, the capacitor will only be 30v when fully charged, and its voltage will drop off proportionally as it discharges - and by V=IR, the current will drop as the capacitor discharges. Making the assumption that the load is purely resistive:
[latex]Vr(t)=V*exp(-t/RC)[/latex]
where Vr(t) is the resistor voltage vs. time, and V is the initial capacitor voltage.

Since a coil is NOT going to be a purely resistive load, but rather a big inductor, the actual result will be somewhat different.
If you really want a quick-and-dirty way to visualize what's going to happen, put it in some circuit simulation software and run a transient simulation. Doing the math to find a mathematical function for the voltage/current vs. time with the inductance accounted for is also not very difficult.
 
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ltspice is free and easy to use for this kind of simple simulation. You can get it from Linear techs website. I have 'simulated' answers to more questions I've had than I've ever asked on this or any other forum =)
 
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