"William E. Sabin" wrote in message
news:9rSbe.21106$WI3.17494@attbi_s71...

Assume the shorting bar is perfect (lossless). Then:

The equivalent circuit just for the coil alone is the used part with its
resistance, magnetically coupled to the shorted part with its resistance.
The coupling from the shorted part to the used part adds an inductive
reactance plus some resistance in series with the used part. The shorted
turns have nearly the same ratio of inductive reactance per inch of coil
length to resistance per inch of coil length as the rest of the coil, so
the Q should not be degraded by this coupling. In other words the net
loss is the same as that of the entire coil operating alone. The perfect
short does not add power loss into the coil.

Assume the shorting bar is not perfect (not lossless). Then:

Some additional power loss is added to the shorting bar and the Q should
decrease.

Bill W0IYH

A followup to my previous input:

I connected an old roller coil with short circuited turns to my ancient
Boonton 260A Q meter.

Over most of the inductor range at 10 MHz the Q remained "fairly" constant.
At the low values of inductance the Q (about 250) dropped about 15%. I
suspect changes in shape factor of the unshorted turns are involved. The
winding pitch of the coil is greatly reduced at low L values to try to
maintain shape factor and improve Q. At low L values, inreasing the
frequency to 15 MHz helped to restore most of the Q.

The detailed analysis of all that is going on in the roller coil is a
time-consuming chore that I don't want to get into. I believe my previous
explanation is OK under ideal textbook conditions (Kraus "Electromagnetics"
4th edition page 235, Eq 2) but of course that is just an approximation to
the actual coil, as could be expected.

On to other things.

Bill W0IYH