# Capacitor charge rate

```cc: tesla-at-grendel.objinc-dot-com

Richard,

In a recent communication, you said a 12kv 120ma 60 cycle power supply will
doing some math and I came up with a different answer.  Maybe others will
check this out and comment.

I calculate as follows:  The available time to charge the capacitor is one
quarter of the 60 cycle waveform (peak voltage).  This is .0042 seconds.  To
achieve 97% charge on a capacitor we need 3 time constants.  The thevenan
equivalent of a 12kv 120 ma transformer must look like an infinite current
12kv supply with a 100k ohm resister in series.  We know we must achieve full
charge (97%) in .0042 sec and it takes 3 time constants to do this so we have
.0042 / 3 = .0014 sec for one time constant. We know one time constant = RC
and C = one time constant / R.  Solving for C we have .0014 sec / 100k ohms =
.014mfd.  Thus, a 12kv 120ma transformer will fully (97%) charge up to a .014
mfd capacitor.

If this math is correct, I calculate that we need a 12kv 171ma transformer to
fully charge a .026mfd capacitor.

What happens when you have a power supply large enough to charge the tank
capacitor in say 1/8 of a cycle?  Then we need a rotary spark gap to take
advantage of the extra power -- do I have this correct?

Ed Sonderman

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