The correct internal-inductance of Windfeld cart.?

Hi Dave,
uff, let me try this. You are right that a higher output voltage (at the secondary) would come at the cost of a lower current --- IF you would not be able to DRAW more current through the cart coil, but you can. That is the WHOLE point.

As to the 47 ohm 'natural impedance' (on primary), I though to have said impedance! Sorry that was a typo, need an editor, methinks. Thanks for pointing it out.

The 10.18 ohms is also impedance since it is in parallel with the reflected secondary impedance of 47 ohm on primary, calculation as we know.

The next item is explained also by drawing more current, as I said on top. You are going into a 10.18 ohm impedance rather than some 100 to 47k! Take your pick, what ever your 'normal' loading would have been.

A cart is actually a 'floating' device, XLR floating, NO ground. So NOTHING is shunted to ground only +/- at least in one of the better scenarios.

>>> The other extreme of this is a situation where a 10 ohm resistor is placed on the primary side of the transformer and the secondary is left open. Then all of the current generated by the cartridge will traverse the 10 ohm resistor and virtually no current will traverse the transformer <<<
That is right on the money, next to nothing can then go through 47k pre-input-impedance...

Yes, and all the parasitic etc., of course but that is actually for the practical reason negligible.

The main point you are missing here is that 20 times the voltage i.e. 6.8mV instead of 0.3mV into 47k pre-input-impedance, WILL and DOES 'pull' more current on the primary side which has ONLY 10 ohm impedance! A virtual short circuit!!

Well, I could go on but let's see if that makes any sense now.
Greetings,
Axel

One more thing to help understand this concept: A cart is a GENERATOR! and not some fixed current source.
Axel

The main point you are missing here is that 20 times the voltage i.e. 6.8mV instead of 0.3mV into 47k pre-input-impedance, WILL and DOES 'pull' more current on the primary side which has ONLY 10 ohm impedance! A virtual short circuit!!

I'm sorry, No it doesn't. In order to not break the first law of thermodynamics, applying the load to the primary of the transformer while keeping a fixed load on the cartridge must DECREASE the current through the transformer.

Furthermore, If you want a 10 ohm load on the cartridge, you can do one of three things,

-10,000 ohms on the secondary of a 31.6:1
-13 ohms on the primary and 47K on the secondary
-dispense with the SUT altogether and place 13R in parallel with the 47K input resistor.

In all three of these situations will draw the same current from the cartridge.

From the transformer perspective, the secondary load will draw more current through the transformer.

the total "source impedance" for whatever is downstream of the 47K resistor is (assuming a 2.5R cart):

Primary load has a source impedance of 2000 ohms
secondary load has a source Z of 2000 ohms
the No SUT has a souce Z of 2 ohms.

dave

OK Dave,
if you would have ever worked with an SUT you'd give me less of a hard time.

I might misread what you try to tell me -- but it sounds like: bumble bees can't fly i.e. must be due also to the: "first law of thermodynamics"...

Please forgive my lack of stamina with all this, at least nobody can say we haven't tried. I will also not ask for a job as physics teacher either, now that I have so miserably failed with such a basic matter as a step-up transformer.

I feel like Galileo must have during inquisition trial: and say it still works...

I'm out of thermodynamics right now.
Greetings,
Axel
PS: your missing thermodynamics come from the rotation of the platter, that imparts energy into the cart, that imparts energy into the coil, that imparts.... don't go ask me now about why the platter is turning, please.

Thanks for your input, Dave.