Sam E brought next idea :
On 12/26/2016 01:19 PM, FromTheRafters wrote:

[snip]

Yet you still seem to think that Ohm's Law applies to open circuits.
What is the 'voltage drop' across the open contacts of such a switch?

It's the supply voltage, unless there's another open somewhere. Current is 0
(so power is 0 too).

P = IE

0 = 0 * 120V

With an ON switch, E=0 and (for example 1A current)

0 = 1A * 0

0W for the switch, whether on or off. It's different if it could be part way
(as in a rheostat for a dimmer).

True about the voltage being the supply voltage, but that is not
'voltage drop' because, as you mention, there is no power (energy)
being dissipated in an open circuit.