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Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Hi If you have a 12v cap and a 450v cap in series, with 400v dc power supply would the 12v cap be destroyed. If the 12v cap was replaced with a spark cap 12kv and a 45kv cap, with a 40kv dc power supply would the spark gap arc?
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
It depends on the capacitances (if we neglect leakage currents, which we can do for a short time, but not for long).
If both caps are equal, they will charge at the same rate, the same volts per Ampere.second passed. So the low voltage one will quickly exceed its rating. In order to limit the voltage to 12v with 400v applied, it would have to have a capacitance of at least 388/12 times that of the high voltage one.
If you apply 40kV to a gap in series with an uncharged capacitor, then you will have 40kV across the gap, at least initially.
Registered Member #3271
Joined: Mon Oct 04 2010, 02:29AM
Location: Canada
Posts: 159
You did not say if the caps both have the same capacitance (DC) or constructions/chemistry (AC).
Have a look at in theory:
Voltage established across each depends on the capacitance.
As far as a spark gap it looks like an open circuit but once the plasma is established it looks almost like a nonlinear short and reactance for a short time. The full AC transient current, voltage, RF oscillations will be across the other poor fellow.
Real caps will also have some leakage current as mentioned by Dr. Slack also and different AC behaviour.
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Thanks you two. If I have a 1uf cap parallel with the spark gap, and 0.001uf cap in series, the spark gap should see almost all the voltage, but what would happen if it arcs and the other 0.001uf cap can handle the voltage, would it arc a second time, before the cap self drains, is it correct the sparkgap would have close to the full voltage drop after the first arc?
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
Andy wrote ...
1uf/(1uf+0.001uf)*40kv = 39960 voltage drop 0.001uf/(1uf+0.001uf)*40kv = 39v
The smaller voltage drop is charged to a high voltage?
Wrong formula. You made it look like a resistive voltage divider. That would, in fact, work if you used 1/C in place of each C.
But you ought to learn why. Do what Dr Slack said, and figure the voltage on each capacitor after 1 microcoulomb flows through both. Hint: then the charge "stored" in each capacitor will be 1 uC.
Andy, you will get to the end faster if you slow down and do less bouncing off the walls. One stepping stone at a time -- save the great leaps until you really know your stuff. I was going to say something about modern medications, but that discussion needn't be in the public forum.
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Different voltage caps in series
