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