Re: Bleed Resistor for Homemade/Large Caps - THE FULL DESIGN NOTE S

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

Original poster: "Black Moon by way of Terry Fritz <twftesla-at-qwest-dot-net>" <black_moons-at-hotmail-dot-com>

1/2 watt will work in series.
5 watt will be needed in par if there 10mohm each(of course, to get the 
same resistance in par, each resistor would need to be much higher 
resistance, so the disipation would lower anyway, as 15 10mohm in par is 
less then 1 mohm) meaning for 10 resistors in series OR par
1/2 watt will work, if the resulting value = 150mohm

>From: "Tesla list" <tesla-at-pupman-dot-com>
>To: tesla-at-pupman-dot-com
>Subject: Re: Bleed Resistor for Homemade/Large Caps - THE FULL DESIGN
>NOTE S
>Date: Tue, 29 Oct 2002 17:54:00 -0700
>
>Original poster: "Jeremy Scott by way of Terry Fritz <twftesla-at-qwest-dot-net>" 
><supertux1-at-yahoo-dot-com>
>
>Okay,
>
>I've had one person say that stringing
>10Mohm 1/2 Watt resistors would work...
>and another saying that they all have
>to be 5 Watts ...
>
>So which is it?
>
>As I understand it a resistor dissipates
>energy as heat. The larger the resistor,
>the more heat it can dissipate due to
>increased contact with the air. Thus, a
>resistor's length and cross section has
>a direct effect on how much power it can
>handle.
>
>So stringing together 15 10Mohm 1/2 Watt
>resistors *should* give 15 times more
>surface area than a single 1/2 Watt resistor.
>This *should* then be able to handle 7.5 watts,
>providing all of the resistors are the same value.
>