DIYbanter - View Single Post

VWWall[_2_]

wrote:
In alt.engineering.electrical VWWall wrote:

| Actually, a real current will flow until the line's capacitance is
| charged to the source voltage. When the source is removed, the energy
| involved will remain until it is leaked off through the inter-wire
| resistance. If the source is AC, no net energy will "flow", except that
| lost in the inter-wire resistance. If the line length is long enough at
| the frequency involved, reflections from the end of an incorrectly
| terminated transmission line will return to dissipate energy in the
| source resistance.

That reflection even happens with DC. When the switch closes, you have a
rising wavefront leading the chargeup of the line. Unless the far end has
a perfectly matched load, that wavefront will reflect back. This is in
fact how a lot of very early radio transmissions were tuned, with the
"switch" being a noisy spark gap, and the "line" being a long wire antenna
cut to a specific length. You don't even need to have 2 conductors.

That's because a switch closure is not really DC. Resolve a step
function into a Fourier series, and it has an infinite number of AC
components. In the case of a single wire, you do need to consider EM
theory.

--
Virg Wall, P.E.
K6EVE
P1-11-49835
T2-11-49835