On Mon, 30 May 2016 16:40:12 -0400, FromTheRafters
wrote:


E=IR
I=E/R
R=E/I

If you state that either I or R is zero (as is the case with open
circuit, which is not really a circuit, or zero resistance attached to
an ideal source)

An open circuit does not have 0 ohms. and the only time you device by
current is to determine resistance knowing only voltage and amperage -
which only works if you have current flow - which means I does not
equal 0. - and an "ideal source" feeding 0 ohms gives you an answer of
infinite current.

the formula doesn't work. There was mention of "zero"
current and in the 'superconductor' comment was about zero resistance.
My analogy with the trains was a bit of fun, but still illustrates that
dividing by zero yields the result of 'undefined' *not* 'zero'.

Undefined or infinite - the current drawn from an "ideal supply" into
0 ohms would be undefined or infinite for a split second untill the
resistance would change due to the heating effect of the current and a
split second later the resistance would become infinite and the
current zero as the "fuse" opened.

I, of course, agree about 'approaching' these limits as being worthy of
consideration. However, that was *not* what I was replying to.

Also, there is no such thing as 'voltage drop' when there is no current
because 'voltage drop' is defined by there being current. You can have
voltage drop across a fuse just before it blows, but the voltage there
after it blows is not 'voltage drop' because there is no current.

Technically trhe voltage drop across an open fuse is considered to be
supply voltage. If you have a series circuit with zero current on all
the defined resistances and therefore no voltage drop across any of
them the total supply voltage is dropped across the "infinite" or
"undefined" resistance element (talking DC) In an AC circuit there
will be a capacitance between the 2 terminals that can be measured,
and the capacitive reactance will cause a miniscule but measurable
current flow