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FromTheRafters

trader_4 explained :
On Tuesday, December 27, 2016 at 2:37:17 PM UTC-5, FromTheRafters wrote:
explained on 12/27/2016 :
On Tue, 27 Dec 2016 09:07:14 -0500, FromTheRafters
wrote:

was thinking very hard :
On Mon, 26 Dec 2016 19:08:42 -0500, FromTheRafters
wrote:

brought next idea :
On Mon, 26 Dec 2016 14:19:03 -0500, FromTheRafters
wrote:

What is the 'voltage drop' across the open contacts of such a switch?

The same as the max voltage supplied at the source.

That is not 'voltage drop', but is the answer I expected from trader_4.
Voltage drop is related to the energy dissipated primarily (but not
exclusively) through the heat created by the current 'flowing' through
the resistance. With no current 'flowing' there is no 'voltage drop' at
all. What you measure there across the open is the supply voltage.
Perhaps more correctly, you measure the voltage drop across the meter's
internal resistance when the meter 'completes' the circuit.

If current were 'flowing' and the wire had resistance *that* would be
'voltage drop'. Only when current is 'flowing', do wires have
resistance and dissipate energy.

A poster going by the nym Al Gebra suggested that Ohm's Law stated that
since the current is zero the voltage must be zero using the form V=IR
and I disagreed. Then trader+4, yourself and IIRC Clare agreed with Al
Gebra at that time. Now you seem to be saying that that isn't so.

Is the voltage zero as suggested by Al, or the same as the supply
voltage? Pick one.

As a thought experiment, consider a length of relatively thin wire
being monitored by an infrared sensor. Knowing other parameters, you
can deduce the 'voltage drop' by the heat being given off (no need to
complete the circuit with a meter). If the current gets too great the
resistance increases and the wire dissipates more and gives off more
heat and the voltage drop increases. Then, eventually, the wire opens,
and you no longer have any 'voltage drop' because there is no current,
but you *do* have supply voltage across the open.

You have 100% voltage drop in an open circuit. It is all just
semantics.

Dismissing the fact that words have meaning won't help you here.

Ohm and the power formulas still work just fine. Resistance is
infinity, Current is zero, power (watts) is zero.

With no power, you have no voltage drop. This is because voltage drop
refers to the energy lost mainly through dissipation which happens when
there *is* charge flowing through a resistance.

No current, no voltage drop.

If you look at a load being 0 ohms to infinity in the theoretical
world of mathematics the analogy works just fine and so does the math.
In reality on the micro scale there probably is some current flow
unless you are doing this in a vacuum and there is no capacitive
coupling. You do not change the rules just because the switch is open
or the wire is broken. That is particularly true if you are a real
world trouble shooter.

The first time this discussion came about, it was about a statement
about Ohm's law that I said IMO was wrong. Sure, I *could* have used
the argument that reality and theory were not the same thing and the
statement about zero amps was already wrong on the face of it. This
*is* about circuit theory, and *not* about reality.

Okay, consider a voltage source charging a capacitor. Consider
everything to be ideal except for the conductors. Only the conductors
can dissipate power, and only in the form of heat. After first applying
the voltage across the cap, a lot of current flows through the
conductors (we can, by convention, say that current flows through a
conductor even though it is not strictly true) and the resistance of
the conductors causes power to be dissipated as heat. The load device
(the cap) has less than the source voltage across it at this time
because the total of the source voltage has been lessened by the amount
of voltage drop across the conductors.

This lost power can be deduced from the total 'voltage drop' across the
conductors' resistance or as the difference between the source voltage
and the voltage at the device being driven (the cap in this case). This
is the voltage drop due to the resistance of the conductors and the
current through them. When the current is zero, that is the cap is
fully charged (in theory it can be, but as you say in practice it can
only be very very close) the loss through dissipation is also zero (or
very very close) -- *not* suddenly as high as the supply voltage as it
seems you are suggesting.

At this time (theoretically) there is no longer any current anywhere in
the circuit, and no 'voltage drop' anywhere, only the supplied voltage
from the source across the cap. No dissipation, no voltage drop. It is
equivalent to two ideal voltage sources connected to each other, since
there is no current, the conductors might as well be ideal.

Basically, this is the same scenario as before, except I have
substituted a charging cap for an opening switch, the difference being
that now we can see the trend of current and voltage drop as they both
approach zero rather than just a sudden change.

And Ohms Law still applies and works. V = IR. I=0, gives V =0,
the voltage drop across the conductors is zero. Nor was there any
division by zero, so don't start in with that again, please.

I wouldn't dream of it, but given I=0 and nothing else, you can't nail
down V nor can you nail down R by using Ohm's Law. Ohm's Law does *not*
tell you that if the current is zero, the voltage must also be zero as
that Al Gebra poster suggested.

Ohm's Law is not only about the V=IR equation, it is about a formula
describing the relationships between these values in a circuit and
requires the other equation forms too. They are derived with the
stipulation that I and R cannot be . . . well, you told me not to go
there, so I won't.