On Monday, May 30, 2016 at 5:59:52 PM UTC-4, Sam E wrote:
Sam, maybe you can explain it to math challenged Rafters. He says
that you can't do what you just did with that simplest of equations
because with a current of zero, you divided by zero. I don't see
any division there, neither do you. Go figure.
I have zero problems with dividing by no.
But with 'voltage drop' you can't have zero current, because 'voltage
drop' is all about the energy delivered to and dissipated by the
device, not the capability of the source to deliver voltage.
A stepper contactor relay's burnt contacts do not have a voltage drop
unless there is current flowing through them no matter how much voltage
the source can deliver to those closed contacts. There is no voltage
drop between the poles of a car battery, unless there is current
through it and some internal resistance to dissipate some of the
energy. Voltage drop is not a static thing like voltage is.
On Mon, 30 May 2016 21:08:12 -0400, FromTheRafters
"voltage drop" testing of a circuit indicates an open circuit by
reading source voltage across the open connection.
A full source voltage drop across a "load" indicates zero current
flow if the resistance is infinite and infinite current if resistance
is zero - with the non- zero and non-infinite values between being
calculatable using ohm's law.
email@example.com submitted this idea :
There's no doubt that you can measure the source voltage across an
opening, that is not the issue here. However, a "voltage drop" is due
to dissipation of energy which doesn't happen in an open circuit (which
isn't even actually a circuit at all).
Right, for the non-zero and non-infinite values - which we aren't
discussing here. However, for the zero current condition, there is no
"voltage drop" at all. The cumulative "voltage drops" in a *closed*
circuit must equal the source voltage - but they are *not* the same
thing. Source voltage does not require current to be flowing, but
'voltage drop' does.
On Tuesday, May 31, 2016 at 7:00:17 AM UTC-4, FromTheRafters wrote:
You're way in over your head here. Obviously you've never passed
algebra or physics 101. From Newton's Laws, we can calculate the
velocity of a ball that is thrown straight up in the air at an
initial velocity of 20m/sec. We can calculate it's velocity at
every point in time. At the peak, Newton's Law gives a velocity
of zero. Are you going to tell us that value of zero has no meaning.
Of course it does, it means the ball isn't moving. Just like a
voltage drop of zero means there is no voltage loss. Capiche?
And a voltage drop of ZERO from Ohm;s LAw with a current of Zero,
says that. Good grief.
On Monday, May 30, 2016 at 9:08:20 PM UTC-4, FromTheRafters wrote:
Neither Sam nor I ever said it was. We said that V = IR. With a current
of zero, V, the voltage drop is zero. YOU are the one that claims that
has "no meaning". I have 3 apples, I take 3 away, how many are left?
Aplles to start - apples taken away = apples left
3 - 3 = 0
That zero, in your world, apparently has no meaning either. It's really
sad, the state of education in America. Even worse when someone so
ignorant has the nerve to try to explain math and science to those of
us that understand it.
On Tuesday, May 31, 2016 at 11:23:43 AM UTC-4, FromTheRafters wrote:
Only to the village idiot who can't pass a trivial 7th grade math question.
V = IR. R=.16, I= 0
What is V?
Our answer, the correct answer even a grade school student knows, is ZERO.
Tell us your answer?
More silly duplicity.
BTW, did you plot that graph for Ohm's Law, of Voltage versus Current?
It's a straight line, that goes right through the origin. We all see
a solid line. You claim there is some unique singularity at the origin,
where there is hole in the line. There isn't.
On Mon, 30 May 2016 16:46:39 -0400, FromTheRafters
Correct - the voltage drop across the open circuit is the supply
voltage.. There is no voltage drop across the conductors, but the sum
of all voltage drops in a circuit MUST equal the supply voltage.
Putting a voltmeter across any segment of a circuit will give the
voltage drop across it. In an open circuit you will read zero except
across the source and across the infinite resistance of an "open
switch" - and both of those will be identical on a DC circuit - and
close enough to identical as to be virtually impossible to measure the
difference on an AC circuit below radio frequencies, where the
capacitance of the :open switch: starts to have a small but
No. they are two separate things. The supply voltage can exist without
any current flow, and the 'voltage drop' cannot. Voltage drop is
because of energy dissipation in a device with current flowing through
Bullshit. If there is no current flowing, there is no voltage drop.
Volts (dropped) = Amps x Resistance. If there are no Amps, it doesn't
matter what the resistance is, volts (dropped) is still zero.
You might argue that when you actually measure the volts at the far
end, you are loading the circuit with your meter but a digital meter
has in impedance in the meg ohms so the current is still virtually
zero and you will still see the full voltage within the accuracy of
As a sanity check, notice voltage drop charts always base the number
on the load in amps times the resistance of the wire.
On Monday, May 30, 2016 at 2:53:56 AM UTC-4, Diesel wrote:
No one ever said it was.
It takes a little
There is no "pushing" with no load.
That's true, but with no load, there is no current, no pushing, no transit.
What we've written is true, it's electricity 101 and quite frankly
if you don't understand these simple basics, you shouldn't be here
As long as the wire has resistance
Sure, as long as there is a load and current flowing. The voltage drop on
each conductor will be V = I * RW, where I is the current flowing and RW is
the resistance of the wire. Set I=0 and what do you get for V?
We have various ways in
Resistors don't change to make up the difference. Go look at the specs
for some heating elements rated for dual voltages. The heater in my
spa for example, is rated at 1500W at 120V, 6000W at 240V. At lower
voltages, they pull less current, not more.
You are out of your area of expertise. With no current flow there is
NO voltage loss. Simple Ohm's Law. Nobody has been found crooked
enough to break that law.
Now, TECHNICALLY you are almost right. If you have a very low
impedence volt meter it will draw milliamps. Say it draws 5 milliamps
(0.005 amps) and the resistance of your cable is 100 ft of #10 copper
which means 200 feet of conductor. At 0..999 ohms per 1000 ft trhat
is 0.1998 ohms. Plug that into Ohm's law -E=IxR - and you get a
voltage drop of (.005 x .1998 )= a whopping 0.000999 volt drop across
the wire. Lets jump out on a limb and say you have a really rotten
voltmeter that draws 50 miliams (.050 amp) and do the math -
That voltage drop will , in the real world, get "lost in the noise" -
well within the accuracy variance of all but the most expensive lab
type volt meters.
Yes, exactly. You are forgetting the "R" in E=IxR.
The starter draws a lot of current because it has a very low
resistance. A 12 volt starter drawing 120 amps has an effectice
resistance of 12/120= 0.1 ohms.and consumes 12X120= 1440 watts -
roughly 1 1/2 hp.
The taser is abour 26 watts. The open circuit voltage is about 50,000
vots. That means the current is (26/50,000)=0.00052 amps.. This means
the impedence or resistance of the tazer is something in the
neighbourhood of (50,000/.00052) = 96,153,846 ohms.
The water theory of electrical theory is a very simplistic
explanation that "doesn't exactly hold water"
firstname.lastname@example.org wrote on 5/30/2016 :
It almost looks like you are agreeing with me now, except you said
'voltage loss' instead of 'voltage drop' which are *not* the same
Also, in another post, you started writing about semiconductors and I
am familiar with forward voltage drop, and it requires a current.
Excerpted from Wikipedia:
"In a small silicon diode operating at its rated currents, the voltage
drop is about 0.6 to 0.7 volts."
Notice the word "currents" in there? There is no 'voltage drop' when no
current is present.
I'm still waiting for "voltage drop" with no reference to current. You
see, devices don't dissipate power when there is no current through
them, so how can there be any 'voltage drop' with no current?
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