Wiring electric baseboard

[FromTheRafters]

trader_4 pretended :

Not if you use the D=RT formula, it is 'undefined' when either R or T is zero (I did *not* say approaching zero).

That is true, unless you state that I or R is zero. See above and below.

Simple math, you are *not* allowed to divide by zero - it is

*undefined*. Infinity is okay to work with because things in that formula can be infinitely small or infinitely large.

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) 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'.

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.

[trader_4]

(I^2)R is the power drop. The voltage drop is IR

And no, it's not about a simple mistake, Diesel doesn't understand Ohm's Law and electricity 101.

[FromTheRafters]

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.

[FromTheRafters]

Power drop?

If there is such a thing in this context, which I doubt, it probably requires current too.

[Mark Lloyd]

[snip]

With any 2 nonzero numbers, A/B is the reciprocal of B/A. In other words, (A/B)*(B/A)=1. If you make A=0, then A/B=0. B/A should then give the reciprocal of 0 (as in, what do you multiply 0 by to get 1). How about infinity?

Still doesn't explain 0/0.

If you don't like this, please don't read it :-)

[clare]

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.

The water theory of electrical theory is a very simplistic explanation that "doesn't exactly hold water"

[hah]

[snip]

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 circuit design.

[clare]

You are not smart enough to brak ohm's law - and the water theory of electricity doesn't exactly hold water.

[clare]

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.

[clare]

You are over your head -- WAY over.

[clare]

I've just posted it for you.

[FromTheRafters]

Maybe you could multiply it by itself zero times. Anything to the zeroeth power equals one.

Okay, so multiplicative groups kick out the zero, it was worth a try.

[gfretwell]

Lost power in the circuit is a function of the voltage drop.

It is all moot in the case of the resistance heater in the example that started this nonsense thread. All of the power lost to voltage drop will still be returned to the home in the form of heat and that was the object of the exercise in the first place.

[FromTheRafters]

[e-mail] brought next idea :

We're not talking about completing a circuit with a meter either are we? Show me how a semiconductor has a voltage drop without any current flowing and maybe I'll take your comments seriously.

[FromTheRafters]

[e-mail] pretended :

Says you. Do you divide by zero for a living?

[FromTheRafters]

Finally, someone who knows what they're talking about. LOL

[Mr Macaw]

Why? Why not just do the sums and work out what's actually needed instead of believing some paper pushing f****it?

[FromTheRafters]

[e-mail] has brought this to us :

Where?

[Sam E]

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.

[FromTheRafters]

[e-mail] 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?