Page 8 of 10
On 05/30/2016 03:41 PM, trader_4 wrote:

I discovered that mistake soon after posting. It is IR.

Most likely, considering the multiple errors (including using the wattage of a heater that wasn't connected).

Sam E expressed precisely :

Power is E***I and since E=I***R power also equals I***I***R or I squared R.

You are correct that there is***no*** 'voltage drop' with no current,
which is not the same as saying that the 'voltage drop' is zero.

#### Site Timeline

- posted on May 30, 2016, 10:04 pm

I discovered that mistake soon after posting. It is IR.

Most likely, considering the multiple errors (including using the wattage of a heater that wasn't connected).

- posted on May 30, 2016, 8:46 pm

Power is E

You are correct that there is

- posted on May 30, 2016, 9:59 pm

On 05/30/2016 03:46 PM, FromTheRafters wrote:

Sorry for the error. I never claimed to be perfect. Although, it'd be hard to make a mistake like that one (voltage drop with no current).

BTW, as to formulae I always liked P=IE.

While 'no' and 'zero' are different ideas, either should be correct here.

Voltage drop = I***R = 0v***R = 0v.

Sorry for the error. I never claimed to be perfect. Although, it'd be hard to make a mistake like that one (voltage drop with no current).

BTW, as to formulae I always liked P=IE.

While 'no' and 'zero' are different ideas, either should be correct here.

Voltage drop = I

- posted on May 30, 2016, 11:22 pm

On Monday, May 30, 2016 at 5:59:52 PM UTC-4, Sam E wrote:

Bingo! 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.

Bingo! 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.

- posted on May 31, 2016, 1:08 am

Sam E brought next idea :

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.

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.

- posted on May 31, 2016, 3:57 am

On Mon, 30 May 2016 21:08:12 -0400, FromTheRafters

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.

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.

- posted on May 31, 2016, 11:00 am

snipped-for-privacy@snyder.on.ca 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.

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

- posted on May 31, 2016, 3:49 pm

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.

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.

- posted on May 31, 2016, 2:00 pm

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.

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.

- posted on May 31, 2016, 3:23 pm

trader_4 pretended :

Which is still wrong no matter how many times you say it.

All of them. Just because you take them away doesn't mean they don't exist anymore. They still exist at the place you took them away to.

Ah, more math, and correct this time - you're learning after all, albeit s-l-o-w-l-y.

It sure is, LOL.

Which is still wrong no matter how many times you say it.

All of them. Just because you take them away doesn't mean they don't exist anymore. They still exist at the place you took them away to.

Ah, more math, and correct this time - you're learning after all, albeit s-l-o-w-l-y.

It sure is, LOL.

- posted on May 31, 2016, 6:30 pm

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.

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.

- posted on May 31, 2016, 3:02 am

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 measureable effect.

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 measureable effect.

- posted on May 31, 2016, 11:05 am

snipped-for-privacy@snyder.on.ca wrote :

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 it.

Equal, yes, but the same thing, no.

[snipped]

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 it.

Equal, yes, but the same thing, no.

[snipped]

- posted on May 31, 2016, 1:55 am

Mon, 30 May 2016

***sigh*** you are correct. I wasted a significant amount of time by
ignoring the fact that the circuit wasn't completed yet. I will eat a
large amount of crow. :)

--

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- posted on May 30, 2016, 3:07 pm

wrote:

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 the meter.

As a sanity check, notice voltage drop charts always base the number on the load in amps times the resistance of the wire.

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 the meter.

As a sanity check, notice voltage drop charts always base the number on the load in amps times the resistance of the wire.

- posted on May 31, 2016, 1:55 am

snipped-for-privacy@aol.com

You are correct. I wasted a significant amount of time by ignoring the fact that the circuit wasn't completed yet. I will eat a large amount of crow. :)

Yes, I've noticed.

You are correct. I wasted a significant amount of time by ignoring the fact that the circuit wasn't completed yet. I will eat a large amount of crow. :)

Yes, I've noticed.

--

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Hmmm. I most certainly don't understand how I can access a copy of a

MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

Hmmm. I most certainly don't understand how I can access a copy of a

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- posted on May 30, 2016, 3:21 pm

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. Capiche?

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 giving advice.

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.

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. Capiche?

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 giving advice.

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.

- posted on May 31, 2016, 1:55 am

Mon, 30
May 2016 15:21:54 GMT in alt.home.repair, wrote:

Aye.

You are correct. I wasted a significant amount of time by ignoring the fact that the circuit wasn't completed yet. I will eat a large amount of crow. :)

Aye.

You are correct. I wasted a significant amount of time by ignoring the fact that the circuit wasn't completed yet. I will eat a large amount of crow. :)

--

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Hmmm. I most certainly don't understand how I can access a copy of a

MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

Hmmm. I most certainly don't understand how I can access a copy of a

Click to see the full signature.

- posted on May 30, 2016, 9:02 pm

wrote:

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 - .050X.1998)=0.00999 volts.

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.

explanation that "doesn't exactly hold water"

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 - .050X.1998)=0.00999 volts.

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.

explanation that "doesn't exactly hold water"

- posted on May 30, 2016, 10:01 pm

snipped-for-privacy@snyder.on.ca 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
thing.

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?

It almost looks like you are agreeing with me now, except you said 'voltage loss' instead of 'voltage drop' which are

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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