Roy Lewallen wrote:
The Rule is that the sum of currents on *all* the box's conductors has
to add to zero. If the box has only two terminals, the sum of the two
has to be zero -- the only way to get around that would be to put Cecil
into the box and have him suck coulombs just as fast as he can. If that
two-terminal box contains an inductor, then the current out has to equal
the current in -- that's the only way the sum of currents at the two
terminals can sum to zero.

That's just the point, Roy. The two terminals don't have to sum to zero
for a distributed network problem, like 1/2WL of coax coiled up inside
that black box.

But one thing you can take to the bank,
folks: the sum of the currents on all the terminals better add to zero.
Unless, of course, Cecil is in the box.

Are you saying I cannot coil up 1/2WL of coax inside a black box and
observe current flowing into both terminals for 1/2 cycle and current
flowing out of both terminals for 1/2 cycle? That would be quite a
revelation.
--
73, Cecil http://www.qsl.net/w5dxp



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You also need to go back and read the fourth sentence of my posting.

Sheesh.

Roy Lewallen, W7EL

Cecil Moore wrote:
. . .
Are you saying I cannot coil up 1/2WL of coax inside a black box and
observe current flowing into both terminals for 1/2 cycle and current
flowing out of both terminals for 1/2 cycle? That would be quite a
revelation.