Why is the neutral tied to ground in USA single phase wiring

[DJ Delorie]

jurb6006@gmail.com writes:
> When that secondary is loaded the impedance of the primary is lower.

If you're drawing real power, then you're creating a load that is at
least partially resistive - this shows up as a slight phase shift in the
relation between voltage and current, both in the primary and the
secondary - which results in a slightly resistive-looking load. If you
don't, the P=IE formula integrates to have no real component, and thus
no real power.

All the hand-waving in the world won't change the laws of physics. You
can mess with the imaginary power all you want[*] (inductive and/or
capacitive) but if you want work done, you need to create some sort of
at least partially-resistive load, which is reflected onto the primary,
which causes losses.


[*] sort of. If you're non-resistive by enough, they ask you to put
compensators at your site to "normalize" the load. Even so-called
imaginary power costs money, because the current through the wires
causes resistive losses.

 

[Nobody]

On Wed, 20 Jul 2016 16:08:29 -0700, jurb6006 wrote:

The transmission line was the primary
of a transformer in series with the load. He created secondary. When that
secondary is loaded the impedance of the primary is lower.

Drawing real power from a cable within the field will create its
own field which induces a voltage in the transmission line. This opposes
the current, requiring work to be done.

There is simply no way that you can reduce either the real or apparent
power in the transmission line by extracting real power from it.

Drawing apparent power (i.e. drawing current 90 degrees out of phase with
the voltage) can either increase or decrease the apparent power of the
transmission line depending upon whether it shifts the phase of the
current toward or away from the phase of the voltage.

But you can't power anything with apparent power.

And any such parasitic power draw is *in addition* to the existing
transmission losses. Adding additional loss doesn't make any existing loss
magically disappear.

> All he did was to pick up some of the loss.

It doesn't work like that. An AC transmission line through the middle of
deep space won't be losing anything. Creating a magnetic field doesn't
(in itself) require any power. Extracting power from that field does.