FW: 60 vs. 30 ma - charging

To: "'Tesla List'" <tesla-at-poodle.pupman-dot-com>
Subject: FW: 60 vs. 30 ma - charging
From: Tesla List <tesla-at-stic-dot-net>
Date: Tue, 24 Jun 1997 13:15:35 -0500
Approved: tesla-at-stic-dot-net

From: 	William Noble[SMTP:William_B_Noble-at-msn-dot-com]
Sent: 	Tuesday, June 24, 1997 1:50 AM
To: 	Tesla List
Subject: 	RE: 60 vs. 30 ma - charging

the voltage on a capacitor is the integral of the current into the capacitor.  
thus, if you hold constant current  the  voltage provides a ramp (dV/dT is a 
constant).  Conversely, the current into a capacitor is the deriviative of 
voltage, eg i=c(dv/dt), so if you put a voltage step from an ideal source into 
a capacitor, you will have a single infinite current spike, thereafter 0.  
Now, of course there is always resistance.

anyway, the terse answer to your question is yes.  However it's misleading.  
If you look at a neon sign transformer as a constant current device, or at 
least as a current limited device, and say that V=Vmax(cos(omega)) then you 
will find that the current into the capacitor is probably at the transformer 
max when omega is near 0, and the current may or may not reach 0 when omega is 
90, depending on the value of C - if the integral of ImaxC for 1/240 of a 
second is > Vmax then the capacitor will charge to Vmax, else it won't 
(remember the voltage goes from 0 to max in 90 degrees = 1/240 sec at 60 hz)

----------
From: 	Tesla List
Sent: 	Monday, June 23, 1997 4:56 PM
To: 	'Tesla List'
Subject: 	Re: 60 vs. 30 ma


From: 	Engle, Daniel (NJAOST)[SMTP:DEngle-at-NJAOST.ML-dot-com]
Sent: 	Monday, June 23, 1997 3:11 PM
To: 	'Tesla List'
Subject: 	RE: 60 vs. 30 ma

I'm not trying to beat a dead-horse here, but I'm trying to see this in
layman's terms.  Feel free to correct me...  If you say that current is
the quantity of electrons(for example, the size of a river-i.e. the
bigger the more water) and voltage is the "pressure"(the speed), then
wouldn't increasing either basically charge the capacitor faster?
Wouldn't doubling the current(increasing the size of the river) or
increasing the voltage(increasing the flow of the water) have the same
effect?

Dan Engle 

	-----Original Message-----
	From:	Tesla List [SMTP:tesla-at-pupman-dot-com]
	Sent:	Sunday, June 22, 1997 9:25 AM
	To:	'Tesla List'
	Subject:	Re: 60 vs. 30 ma


	From: 	Alfred A.
Skrocki[SMTP:alfred.skrocki-at-cybernetworking-dot-com]
	Sent: 	Wednesday, June 18, 1997 7:29 PM
	To: 	Tesla List
	Subject: 	Re: 60 vs. 30 ma

	"You can make all the current in the world available to a
capacitor 
	and it is NOT going to charge any faster! To make a capacitor
charge
	faster you have to increase the applied voltage! Try and
remember the
	old analogies to clearify; current is the quantity of electrons
and
	voltage is the pressure the electrons are under. How any
electrons 
	you have available wont effect how fast a capacitor charges, but
the 
	amount of pressure (voltage) they are under will! If you keep
the 
	applied voltage the same but double the available current, you
will
	have to double the capacitance to use that available current."