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Re: Rewrite of Mutual Inductance Laws for Tesla List
Original poster: "Paul Nicholson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <paul-at-abelian.demon.co.uk>
> L1 =10.8 mh, C1 =14 uf
> L2= 60 henry, C2= 1 nf
> Input = 1.67 volts *.35 A = .58 VAR
> Output=(4/3.16*1000)*.004A = 5.06 VAR
Unfortunately, the jpg doesn't make clear the circuit arrangement, so
we're not sure just where the V and I measurements were picked up.
But it's clear you've got a pair of coupled LC's both resonant near
your alternator frequency.
I suspect that your input measurements are made between the alternator
and L1C1, in which case if L1C1 is tuned to the alternator frequency
these values will represent *real* power, not VAR.
Whereas with the L2C2 measurements, the current can only be the reactive
circulating current (since you've not mentioned a load), and therefore
the 'output' reading is really the VAR of the stored energy in L2C2
On this basis, we can estimate the Q factor of your combined L1C1/L2C2
system. Your 5.06 VAR represents a stored energy of 1680uJ (since this
energy circulates back and forth between L2 and C2 some 2 * pi * 480
times per second).
If 0.58 Watts of real power are necessary to replenish the leakage from
your store of 1680uJ, then the Q is given by
Q = 2 * pi * frequency * stored energy / input power
= 2 * 3.141 * 480 * 1680e-6/0.58
So I'm afraid I don't see anything in your post which indicates a
problem with mutual inductance. At times like this you always have to
ask yourself which is the most likely of two possibilities: Either
you've misunderstood the operation of the circuit, or some very well
tried and tested laws of physics are wrong.