V
VWWall
Guest
P E Schoen wrote:
lights, was an example. There are still lots of these around!
--
Virg Wall
The old "constant current" transformers, used to power series street"Fred Abse" wrote in message
newsan.2013.05.28.19.58.35.835800@invalid.invalid...
Try this: Note the syntax of the last line. The
exclamation point shows it to be a directive.
This example gives 9.118 volts peak at the tap.
Take the 1 milliohm rser out of the voltage source,
the matrix goes singular.
I used the RMS values and I get a ratio of 0.0914, rather than 0.1 as
might be expected. 100 H should be 10 times the number of turns for 1 H,
but because the inductors are coupled it does not quite work that way.
99 H corresponds to 9.95 turns while 1 H is 1 turn. Thus the turns ratio
is 1/10.95 = 0.0913.
Changing the K factor to a comment shows a ratio of 0.01, proving that
non-coupled inductors function as pure reactance and their inductances
may be used for the voltage ratio, just like resistors.
A coupling constant of 0.09 yields a ratio of 0.084.
0.5 shows a ratio of 0.054
0.1 => 0.0196
What sort of transformer might have such a low coupling factor?
Could two isolated inductors be arranged mechanically to have an
adjustable coupling factor, and thus function as an isolated variac?
0 => (what do you think?)
Paul
lights, was an example. There are still lots of these around!
--
Virg Wall