Re: Bleed Resistor for Homemade/Large Caps EXPERIMENTS

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

Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>" <Mddeming-at-aol-dot-com>

In a message dated 10/30/02 10:26:54 AM Eastern Standard Time, 
tesla-at-pupman-dot-com writes:


>Original poster: "Jeremy Scott by way of Terry Fritz <twftesla-at-qwest-dot-net>" 
><supertux1-at-yahoo-dot-com>
>
>Continuing this...
>
>I soldered together a string of 15 10M ohm
>1/2 watt resistors and connected them
>across the leads of the capacitor.
>
>I then connected the leads of the capacitor
>to my 15kV NST and charged it up.
>
>The capacitor is two 35kV Maxwell pulse caps
>in series so I'm sure they can take the abuse
>of a direct NST connection.
>(cap value is larger than resonant so i doubt
>there will be over-voltage issues...)
>
>I charged and discharged the capacitor without
>any lethal residual charge, none of the resistors
>exploded and all remained cool to the touch.
>
>I fear I might have damaged the NST though --
>not sure on this one. The secondary winding
>resistance is still about 6.3K, with 3.15K on
>either side of ground...I believe the secondary
>windings are okay. The resistance of the primary
>is only about 1/2 an ohm! I'm not sure if this
>is right for a properly working 15kV NST...someone
>want to verify? (How could the primary windings
>get killed anyway?)
>
>I decided to use my resistor array to try and measure
>the output voltage. With 15 resistors of all the same
>value and 15Kv, the voltage across each should be
>1000V. attaching the meter across the a resistor, the
>maximum voltage I was able to read was 600VAC
>
>Am I missing something, or does this meant that
>maximum output is now only 9000VAC ?


Hi Jeremy,
        The problem is one of physics: The act of observing changes that 
which is being observed. To measure the actual voltage, the input impedance 
of the meter must be very much greater than the resistance across which the 
measurement is being taken. In this case Z(in)>>10Megohm may not be true. 
Let's say that your meter has Z(in)=10 Megohm. Then WHILE you are measuring 
the voltage across a 10Meg resistor, your act of measurement has made it 
effectively 5 Megohm, because R(eff)=Rres. x Rmeter /(Rres.+Rmeter) and the 
voltage drop will be less by half. If your meter has Z(in)=100 Megohm, then 
10 x 100/(10+100)= 9.09 Meg and the voltage you read will only be low by 
10%, and with 200Meg Z(in) the error introduced by the meter would be less 
than 5%. When measuring very large or very small quantities, one must 
account for the effect of the measuring device itself.
If you are using a pair of caps in series, each should have its own bleeder 
resistor at half the max voltage rating, because if one should die open, 
the other could still be fully charged.
Hope this helps.

Matt D.