On Apr 18, 3:27*am, Roy Lewallen wrote:
Roger Sparks wrote:
. . . But how can we have a source with zero resistance, zero
capacitance, and zero inductance because in the real world, any
source has impedance? *The short has "zero resistance, zero
capacitance, and zero inductance but it does not emit energy nor have
a reverese voltage, both properties of the voltage source. *It is not
reasonable to assign the properties of the short to the voltage
source, ignoring the reverse voltage situation, and expect reflectons
from the source to be identical to reflections from a short.

None of the ideal components we use for linear circuit analysis exist in
the real world. We use ideal resistances which have no inductance or
capacitance, capacitances which have no resistance or inductance, ideal
inductances which have no resistance or capacitance, ideal controlled
sources which operate over an infinite range of control and output
values. And try making anything even vaguely resembling an ideal
transformer. So what's the problem in accepting an ideal voltage source
as another model element? If you want a better approximation of
something you can build in the real world, add an ideal resistance to
the ideal voltage source, and you'll have a much better representation
of most real sources.

As for the way the source reacts to an impinging wave, note that the
voltage across a short circuit doesn't change when a wave hits it.
Neither does the voltage across an ideal voltage source. Consequently,
they do have exactly the same effect on waves.

Roy Lewallen, W7EL

Perhaps it would help clarify the thinking to plot some voltage-
current
curves.

If we plot V versus I for a resistor (with V on the vertical access)
we
get a line with a slope equal to R. This line passes through the
origin.

For a short, the line is horizontal (i.e. slope and R are zero) and
for
an open the line is vertical (i.e. slope and R are infinite).

Now plot the V-I characteristic of a resistor in series with an ideal
voltage source. Again it is a line with a slope equal to R, but it
does
not pass through the origin, it crosses the vertical axis at the
voltage
provided by the source. So the y-axis crossing is controlled by the
voltage source and the slope is controlled by the resistor. If you
reduce the resistor to zero, you get a horizontal line crossing
the y-axis at the voltage of the source. The line being horizontal
means that no amount of current will change the voltage.

We often talk of resistance as V/I, but there are many situations in
which it is better to think of it as deltaV/deltaI (or, in the limit,
dV/dI); that is, the change in voltage that accompanies a change
in current. This is exactly the slope of the V/I curve at that point
and works for computing resistance regardless of whether there is a
voltage offset present.

And this is how the source resistance of a battery or power supply
would be measured. Measure the voltage and current at one load,
change the load, measure the new voltage and current, compute the
source resistance from the changes in the voltage and current.

This dV/dI view of resistance is extremely useful and can be
applied to devices with very complex V-I curves. Consider a
tunnel diode (http://en.wikipedia.org/wiki/Tunnel_diode).
Over part of its V-I curve, the slope (i.e. resistance) is
negative. If we put a tunnel diode in a circuit with appropriate
bias such that the tunnel diode only operates over this range
of its V-I curve, then for the purposes of that circuit it
can be modelled as a resistor with negative resistance.

Think R=dV/dI, not R=V/I.

...Keith