On Mon, 30 May 2016 11:45:57 -0400, FromTheRafters
You are burying yourself deeper.
Quit while you are ahead. If there is no semiconductor in the "open
circuit" there will be no voltage drop. As soon as you put a meter on
to check the voltage it IS a circuit.. A smiconductor has a "forward
voltage drop" that behaves differently than a resistance - but we are
not talking about semiconductor physics here.
On Mon, 30 May 2016 17:26:52 -0400, FromTheRafters
The voltage drop across a perfect semiconductor is not affected by
current - other than requiring the current to be NON zero. - and it is
vityually impossible to read voltage anywhere in a circuit without
completing the circuit - in a DC circuit requiring a resistive load -
an an AC circuit either a resistive, inductive or capacitive load. -
which means you can NOT measure voltage in a totally unloaded or
There you go then. Besides, semiconductors are sometimes incompatible
with Ohm's Law.
True, but that is not the point. Does voltage drop even exist
theoretically when there is theoretically no current?
By definition, you need current to have energy dissipation in the
device and thus a 'voltage drop'. If a device is 'heating up' you can
be sure there is current through it without attaching a multimeter and
completing a previously open circuit (which isn't actually a circuit -
since it is open).
On Monday, May 30, 2016 at 11:11:39 PM UTC-4, email@example.com wrote:
Sigh. Sadly that is incorrect too. The ideal diode equation for the
perfect semiconductor shows the relationship between the current through
the diode and the voltage across it. It's exponential. The current is
a function of the voltage, it's just that the current goes up a lot for
a small change in voltage.
- and it is
Of course you can measure voltage in an open circuit. Stick the
test leads of a high impedance meter into a receptacle with no loads
on that circuit. For all practical purposes, you're measuring the
open circuit voltage, you'll see ~120V. The only load is the insignificant
impedance of the meter, in the millions of ohms. That load is so
small, it's less than the accuracy of the meter itself. And if you want to
use more exotic test equipment, clearly we could measure it without
even that insignificant load.
We have a guy here who doesn't understand basic algebra and Ohm's Law.
IDK why you're dragging in semiconductor physics, AC circuit dynamics
and leading the discussion into rat holes.
It happens that firstname.lastname@example.org formulated :
No, *he* knew quite well what he was talking about.
Ah, engineering, no wonder the math skills are so weak. They use
calculus all day long but few of them know how or why it works.
Show me a definition of "voltage drop" that doesn't involve current
flowing in a circuit. I'll wait, but I haven't got forever so be quick
On 05/30/2016 01:25 PM, FromTheRafters wrote:
Normally, the voltage on a wire will be the same at all points
regardless of wire size or current flow. "Voltage drop" is what happens
when the copper atoms get tired of being so reasonable, and assert their
need for "silly time" (its their version of a "smoke break"). Since they
can do this at any time, it needs to be accounted for in electrical
On Monday, May 30, 2016 at 2:25:40 PM UTC-4, FromTheRafters wrote:
V = IR is the formula for voltage drop. Put in zero for I, any finite
resistance for R, you get V = 0. That tells you there is no voltage drop.
And yes, contrary to your BS, zero does have meaning. Hell, you can
even graph this, V versus I, it's a straight line and it goes right
through the origin. At zero current, the voltage is zero and yes, that
zero has meaning. PS: I didn't do any division by zero either.
On Mon, 30 May 2016 20:13:21 -0400, FromTheRafters
No voltage drop across the wire does not mean no voltage drop across
the circuit thogh.
You've ( it appears) been arguing both sides of the equation.
To avoid confusion - WHOEVER said there would be a voltage loss in an
unloaded wire is WRONG.
Also whoever said there is no voltage drop in an open circuit is ALSO
And to top it all off, in order to measure the voltage across an open
circuit, you MUST close the circuit - meaning it is no longer an
"open" circuit, AND
In the real world there is no such thing as zero ohms. You can get
REAL close - but "in the wild" it does not exist. - so you are never
REALLY deviding by or multiplying by ZERO when solving ohm's law
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