On Mon, 30 May 2016 16:46:39 -0400, 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.

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.


Correct - the voltage drop across the open circuit is the supply
voltage.. There is no voltage drop across the conductors, but the sum
of all voltage drops in a circuit MUST equal the supply voltage.
Putting a voltmeter across any segment of a circuit will give the
voltage drop across it. In an open circuit you will read zero except
across the source and across the infinite resistance of an "open
switch" - and both of those will be identical on a DC circuit - and
close enough to identical as to be virtually impossible to measure the
difference on an AC circuit below radio frequencies, where the
capacitance of the pen switch: starts to have a small but
measureable effect.