RE: MOT arc

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

Original poster: "Loudner, Godfrey by way of Terry Fritz <twftesla-at-qwest-dot-net>" <gloudner-at-SINTE.EDU>

Hi Marc

The expression di/dt denotes the first time derivative of the current, i.e.,
the rate of change of current with respect to time. If one know the wave
shape of i as given by a function of time, then one can calculate di/dt
using differentiation techniques from calculus.
However you don't need to know any calculus to see how the formula V =
-L(di/dt) applies to this MOT situation. When the battery is disconnected
from the MOT primary, the current decreases to zero in an extremely short
period of time. This means that -di/dt is an extremely large number (note
the - sign) for those values of t after the battery disconnection. Hence V
can be very large. For example let L = 20 henry and di/dt = -2000
ampere/second. I am using -2000 ampere/second as a uniform average value.
Then the formula for the inductive voltage gives V = 40,000 volt. An
interpretive scheme is that as the current drops, the inductive voltage
rises to maintain the current. 

Godfrey Loudner 

> -----Original Message-----
> From:	Tesla list [SMTP:tesla-at-pupman-dot-com]
> Sent:	Tuesday, June 19, 2001 6:57 PM
> To:	tesla-at-pupman-dot-com
> Subject:	Re: MOT arc
> 
> Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <MShock8073-at-aol-dot-com>
> 
> In a message dated 6/19/01 10:46:56 AM Eastern Daylight Time, 
> tesla-at-pupman-dot-com writes:
> 
> << Connect the 6 volts to the MOT primary, and then disconnect quickly!! V
> = L
>  x di/dt (L = primary inductance in Henries) tells you that you can get a
>  humongus inductive kick off the primary and a massive voltage off the
>  secondary also. >>
> 
> Can someone explain to me how you get the di/dt values?
> 
> Marc S.
> 
>