There is no cancellation because the postive and negative peaks do not
occur at the same time.


Paul Burridge wrote in message . ..
On Sat, 13 Nov 2004 11:42:11 +0000, Scott
wrote:

Huh? It's only 5:30AM here and I just got up but, the ONLY time you
aren't consuming power is at the zero crossing of the voltage and
current sine waves (assuming a purely resistive load where I and E are
in phase). Since you are paying for power, which is P=I X E, during the
negative half cycle, you have, for example, -168 Volts X -1 Amp = +168
Watts...try it on a calculator...negative times a negative is positive.

Thanks, Scott. So you're basically agreeing with me. I owe the power
co. for the positive cycles they send me; they owe *me* for the
negative ones. Since they are equal and opposite, they cancel each
other out. Overall, then, zero billing justified.
We are being conned!!!

Huh? It's only 5:30AM here and I just got up but, the ONLY time you
aren't consuming power is at the zero crossing of the voltage and
current sine waves (assuming a purely resistive load where I and E are
in phase). Since you are paying for power, which is P=I X E, during the
negative half cycle, you have, for example, -168 Volts X -1 Amp = +168
Watts...try it on a calculator...negative times a negative is positive.

P=I^2R, so which direction the current is flowing is irrelevant, as the
squaring removes any negatives, and R is always positive.

Would be interesting if you could get a true -R though! (not like a tunnel
diode, which just has a small region where increasing V decreases I, but
it's still positive)

On Fri, 19 Nov 2004 08:26:54 -0500, "Dave VanHorn"
wrote:


Huh? It's only 5:30AM here and I just got up but, the ONLY time you
aren't consuming power is at the zero crossing of the voltage and
current sine waves (assuming a purely resistive load where I and E are
in phase). Since you are paying for power, which is P=I X E, during the
negative half cycle, you have, for example, -168 Volts X -1 Amp = +168
Watts...try it on a calculator...negative times a negative is positive.

P=I^2R, so which direction the current is flowing is irrelevant, as the
squaring removes any negatives, and R is always positive.

Would be interesting if you could get a true -R though! (not like a tunnel
diode, which just has a small region where increasing V decreases I, but
it's still positive)

---
Overunity?!

--
John Fields

Would be interesting if you could get a true -R though!

I spent years perfecting a bar magnet with only a North pole. (Monopolar) I set
it down on a table for a moment and it promptly took off for Antarctica. I
haven't seen it since.

I've been making them with only South poles for the last year, but have them
all tied down. I'm going to release them on December 24 and give Santa a big
surprise.

On Fri, 19 Nov 2004 08:26:54 -0500, Dave VanHorn wrote:


Huh? It's only 5:30AM here and I just got up but, the ONLY time you
aren't consuming power is at the zero crossing of the voltage and
current sine waves (assuming a purely resistive load where I and E are
in phase). Since you are paying for power, which is P=I X E, during the
negative half cycle, you have, for example, -168 Volts X -1 Amp = +168
Watts...try it on a calculator...negative times a negative is positive.

P=I^2R, so which direction the current is flowing is irrelevant, as the
squaring removes any negatives, and R is always positive.

Would be interesting if you could get a true -R though! (not like a tunnel
diode, which just has a small region where increasing V decreases I, but
it's still positive)

Instead of wires, use glass tubes full of plasma. ;-)

Cheers!
Rich