Re: 60 vs. 30 ma

To: "'Tesla List'" <tesla-at-poodle.pupman-dot-com>
Subject: Re: 60 vs. 30 ma
From: Tesla List <tesla-at-stic-dot-net>
Date: Tue, 24 Jun 1997 22:17:04 -0500
Approved: tesla-at-stic-dot-net


From: 	Rick Holland[SMTP:rickh-at-ghg-dot-net]
Sent: 	Monday, June 23, 1997 11:13 PM
To: 	Tesla List
Subject: 	Re: 60 vs. 30 ma

Tesla List wrote:
> 
> 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

-- 
Forget rivers. We're talking electronics. The formula for attaining 63%
of maximum charge (ie. applied VOLTAGE) in terms of time is R*C. The
number of time constants required to approximate full charge is 3. In
order to decrease the amount of time necessary to charge the capacitor,
you must decrease R or C or increase voltage. In the event you increase
voltage, the resultant 63% will be that fraction of a higher voltage.
Increasing current and leaving all other factors even will not decrease
charge time or increase charge per time unit.

      Rick Holland

      The Answer is 42