Re: MMC cap bank

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "jimmy hynes by way of Terry Fritz <teslalist-at-qwest-dot-net>" <chunkyboy86-at-yahoo-dot-com>

Hmmm... but how do it know ;-)?
The current flowing *in each burst* is completely
independent (assuming fixed firing voltage) of the
break rate, so how does one burst know to lose more or
less energy? It seems to me that identical
circumstances would lead to identical loss ;-)).

Does the first burst waste a different amount of
energy because it doesn't know the real break rate
;-))?

Why do you say that the RMS current is 10x? I agree
that the RMS is  the "equivalent" AC current to a
fixed DC current. To find the RMS you square, average,
and then take the square root. After squaring and
averaging, the result for 1000 BPS will be 10 times
the answer for 100 BPS, because the squaring affects
them both the same. After taking the square root, the
ratio droppes to SQRT10 right???

BTW- Thanks for being such a cool moderator guy ;-))

--- Tesla list <tesla-at-pupman-dot-com> wrote:
 > Original poster: "Terry Fritz" <teslalist-at-qwest-dot-net>
 >
 > Hi Jimmy,
 >
 > At 01:12 PM 7/4/2003 -0700, you wrote:
 > >Hi Terry,
 > >
 > >I understand the I^2R thing, but isn't the RMS
 > current
 > >only square root of 10 times as much? After you
 > square
 > >and average it, it would be 10 times, but you still
 > >have to take the square root, right?
 >
 > Nope.  RMS is just the "equivalent" AC current to a
 > fixed DC current.  If I
 > run 10X the BPS rate, I draw 10X the current and get
 > 100X the heat loss
 > across a resistor.
 >
 >
 > >If it really is 500 watts lost with 1000 BPS, then
 > the
 > >amount of energy lost per pulse is 0.5 joules. If
 > you
 > >lose 5 watts at 100 BPS, then the energy lost per
 > >pulse is 0.05 joules. What would cause the increase
 > in
 > >loss per break?
 >
 > It's not linear.  "I^2"  The graph is a steep ski
 > slope.
 >
 > >We are talking about the same energy
 > >per pulse right?
 >
 > Yes.  Power = current squared times resistance.
 > Double the current, four
 > times the power.  Triple the current, nine times the
 > power...
 >
 >
 >
 > >--- Tesla list <tesla-at-pupman-dot-com> wrote:
 > > > Original poster: "Terry Fritz"
 > <teslalist-at-qwest-dot-net>
 > > >
 > > > Hi Jimmy,
 > > >
 > > > Suppose we have 10 amps RMS at 100 BPS and our
 > caps
 > > > are 0.05 ohm of
 > > > internal resistance (typecal for a 15/60).  From
 > > > P=I^2R the power lost as
 > > > heat in the caps is 5 watts.  Now lets hook it
 > to a
 > > > pole transformer and
 > > > run it at 1000 BPS for 100 amps RMS:
 > >Use my signature  Allow HTML tags [Preview]
 > >
 > > >
 > > > P = I^2 x R  == 100^2 x 0.05 = 500 watts.
 > > >
 > > >   The array can run all day at 5 watts.  But 500
 > > > watts it will die
 > > > fast!!  Probably like 15 seconds.
 > > >
 > > > Cheers,
 > > >
 > > >          Terry
 > > >
 > > > At 09:07 PM 7/3/2003 -0700, you wrote:
 > > >
 > > > > > Second, keep the BPS rate LOW.  Capacitor
 > > > current is
 > > > > > directionally
 > > > > > proportional to BPS.  3000 BPS has 10X the
 > > > current
 > > > > > of 300 BPS and 100X the
 > > > > > cap heating!!!
 > > > >
 > > > >Hi Terry,
 > > > >
 > > > >I "think" the heating should only be 10x as
 > much.
 > > > If
 > > > >it were 100x then each burst would have to
 > waste
 > > > 10x
 > > > >the power, and I don't see how that is
 > possible.
 > > > The
 > > > >average current will be 10x as much, but the
 > RMS
 > > > will
 > > > >only be SQRT10x as much because the duty cycle
 > is
 > > > >greater. Check me on this because it's getting
 > late
 > > > >for me ;-)).
 > > > >
 > > > >=====
 > > > >Jimmy
 > > >
 > > >
 > >
 > >
 > >
 > >=====
 > >Jimmy
 >
 >



=====
Jimmy

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