Re: electrical power

[Larry Cartwright <exit60@CABLESPEED.COM>]

V and R have to refer to the same system, Rick.  You are using V to
refer to the voltage or potential across the entire distribution system,
but R to refer to the resistance of only the wires.  You have to make a
choice that matches V and R appropriately, such as:

(1) V is the voltage or potential across the source of the entire
distribution system, and R is the total resistance of that system
(wires, loads, etc).

(2) V is the voltage or potential difference between one end of a wire
and the other end, and R is the resistance of that length of wire.

(3) V is the voltage or potential across a load at the customer end of
the distribution network and R is the resistance of that load.

If you take V to be the difference in voltage or potential from one end
of the wire (or other resistance) to the other (i.e. voltage drop), then
P=V*I and P=I^2*R and P=V^2/R all refer to the power which produces the
heating in the wire.

Best wishes,

Larry

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Larry Cartwright   <exit60@cablespeed.com>
Retired (June 2001) Physics Teacher
Charlotte MI 48813 USA
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Rick Tarara wrote:
> ----- Original Message -----
> From: "John S. Denker" <jsd@MONMOUTH.COM>
>
> > Justin Parke wrote:
> > > Does anyone have any notes written
> > > about the distinction between P=VI and P=I^2R?
> >
> > In an ohmic situation, there is no distinction.
> >
> > In a non-ohmic situation, P=VI is exact and
> > I2^R has no physical significance.
> >
> > > I understand that the latter refers to
> > > "joule heating loss" but I don't feel clear as to why.
> >
> > P=IV is the Joule heating.
> >
> > Anything else is an approximation at best.
>
> I guess I don't understand the last part here.  A specific example:  I want
> to supply a small town with electrical power.  I need to supply
> approximately 10 MW.  I could try and do that at 100,000 V and 100 A, or at
> 100 V and 100,000 A.  In both cases P = IV = 10MW.  However, for whatever
> the resistance of the wires delivering the power, I^2R will be quite
> different.
>
> Rick
>
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