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NST Max Ratings and Mains Resonance (fwd)
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From: Alfred C. Erpel [SMTP:aerpel-at-op-dot-net]
Sent: Tuesday, February 24, 1998 6:31 PM
To: 'List, Tesla'
Subject: Re: NST Max Ratings and Mains Resonance (fwd)
*******************
I am replying to this message late because of email problems.
*******************
>From: Malcolm Watts [SMTP:MALCOLM-at-directorate.wnp.ac.nz]
>Sent: Sunday, February 15, 1998 6:50 PM
>To: Tesla List
>Subject: Re: NST Max Ratings and Mains Resonance (fwd)
>
>Hi Alfred,
>
>> Date: Sat, 14 Feb 1998 11:24:38 -0500
>> From: "Alfred C. Erpel" <aerpel-at-op-dot-net>
>> To: Tesla List <tesla-at-pupman-dot-com>
>> Subject: Re: NST Max Ratings and Mains Resonance (fwd)
>>
>
>>
>> >
>> >
>> >---------- Forwarded message ----------
>> >Date: Thu, 12 Feb 1998 15:54:09 -0500
>> >From: Thomas McGahee <tom_mcgahee-at-sigmais-dot-com>
>
>> >Subject: NST Max Ratings and Mains Resonance
>> >
>> >Bill, Malcolm, and other interested coilers,
>> >Besides the resonant rise that Malcolm mentions, there is also
>> >the fact that when the main cap and the transformer are set to
>> >resonate at the mains frequency, the transformer is capable
>> >of providing current levels that are several times the usual
>> >'current-limited' value. If the wire in the secondary is too
>> >thin, then you can actually burn out the secondary winding
>> >under these mains-resonant conditions.
>> >
>> >Thus, with the proper resonant conditions a 15KV 60 MA
>> >NST can charge the mains cap up to voltages in excess of
>> >40KV, and at a rate that is much greater than the 60 MA
>> >rating would suggest. Note that both the extra voltage and
>> >extra current can contribute to the NST failing prematurely.
>> >
>> >Hope this helps.
>> >Fr. Tom McGahee
>>
>>
>
>> Hello,
>>
>> It is my understanding that in a series resonant circuit, the
>> capacitive reactance and inductive reactance exactly cancel out, leaving
>> only the pure resistance (ohms) as the total circuit impedance.
>> Therefore,(at resonance) the current (I) flowing in the circuit is
>> determined by I=E/R where E = input voltage (to the series
>> resonant circuit) and R = ohms resistance of the circuit.
>>
>> My point is, I can see how power (EI) is increased in an inductive
>> circuit because voltage is increased (and of equal value) measured across
>> the capacitor and inductor, but the current (I) flowing in the circuit
has
>> not changed.
>> What resonant conditions allow/cause an increased current flow?
>
>If the voltage across a component (resistive or reactive) increases,
>surely the current through it must have increased if the frequency and
>component values haven't changed?
>
>Malcolm
>
The current flow in a series resonant circuit is determined by the
voltage applied and the pure resistance (not reactance's because they are
cancelled out) I=E/R. The voltage across each reactive component (capacitor
and inductor) is determined by E=IR after having calculated I, above. (R is
the components reactance, and to keep this simple, I am assuming ideal
components with no resistance).
So where is the current increase?
If I am wrong, please show me where.
Regards,
Alfred Erpel