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FromTheRafters

submitted this idea :
On Mon, 30 May 2016 21:08:12 -0400, FromTheRafters
wrote:

Sam E brought next idea :
On 05/30/2016 03:46 PM, FromTheRafters wrote:
Sam E expressed precisely :
On 05/30/2016 06:13 AM, Al Gebra wrote:

[snip]

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be zero
as well.

IIRC, it's I squared R. Of course, it's still no voltage drop with no
current. Is it possible there's a confused poster here, who has R and
thinks it's E.

Power is E*I and since E=I*R power also equals I*I*R or I squared R.

Sorry for the error. I never claimed to be perfect. Although, it'd be hard
to make a mistake like that one (voltage drop with no current).

BTW, as to formulae I always liked P=IE.

You are correct that there is *no* 'voltage drop' with no current, which
is not the same as saying that the 'voltage drop' is zero.

While 'no' and 'zero' are different ideas, either should be correct here.

Voltage drop = I*R = 0v*R = 0v.

I have zero problems with dividing by no.

But with 'voltage drop' you can't have zero current, because 'voltage
drop' is all about the energy delivered to and dissipated by the
device, not the capability of the source to deliver voltage.

A stepper contactor relay's burnt contacts do not have a voltage drop
unless there is current flowing through them no matter how much voltage
the source can deliver to those closed contacts. There is no voltage
drop between the poles of a car battery, unless there is current
through it and some internal resistance to dissipate some of the
energy. Voltage drop is not a static thing like voltage is.
"voltage drop" testing of a circuit indicates an open circuit by
reading source voltage across the open connection.

There's no doubt that you can measure the source voltage across an
opening, that is not the issue here. However, a "voltage drop" is due
to dissipation of energy which doesn't happen in an open circuit (which
isn't even actually a circuit at all).

A full source voltage drop across a "load" indicates zero current
flow if the resistance is infinite and infinite current if resistance
is zero - with the non- zero and non-infinite values between being
calculatable using ohm's law.

Right, for the non-zero and non-infinite values - which we aren't
discussing here. However, for the zero current condition, there is no
"voltage drop" at all. The cumulative "voltage drops" in a *closed*
circuit must equal the source voltage - but they are *not* the same
thing. Source voltage does not require current to be flowing, but
'voltage drop' does.