Wiring electric baseboard

[Sam E]

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).

[Mark Lloyd]

[snip]

There's power dissipated by the wire (mostly as heat). And that does require current.

[clare]

You are WAY in over your head.

[trader_4]

Idiot.

There is no division in Distance Traveled = rate x time. If either the rate or the time is zero, the distance is zero. And obviously it has meaning, it means the train did not move.

See this:

V = IR. There is no division by zero. If I or R is zero, V is zero. And that zero has meaning.

We're not dividing by zero. YOU just keep pretending we are.

If either current or resistance is zero, Voltage is zero and contrary to your BS, it has meaning.

Is there zero resistance in that heater circuit wire?

Anyone here doing that division, with a current of zero, to try to calculate resistance? No.

V = IR If either I or R is zero, V is zero. Did you even take algebra?

That right, because with no load, you have zero current. Put zero in for I above, put a finite value for R and you have zero voltage.

That was brought up by the guy who doesn't even understand Ohm's Law. No point in going there, it has nothing to do with the current discussion.

I does nothing of the sort. All it demonstrates is that:

Distance Traveled = Rate x Time.

Rate of zero, Distance Traveled is zero. It has meaning the train didn't move. Idiot.

Idiot. If a fuse blows, the voltage across it after it blows is the full open circuit voltage. Try using a meter and see.

[trader_4]

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.

[trader_4]

Yeah, here's the Rafters guy trying to explain electricity basics and algebra to us, and he can't even grasp the idea of the power drop that corresponds to the voltage drop.

[trader_4]

He's right, you're in way over your head. This silliness over Ohm's law started over the voltage drop over a 100 ft of #12 wire. It's also referred to as voltage loss. Everyone here in the thread at least understands that.

And again, V= IR. With a current of 14 amps, a wire resistance of .16 ohms you get a voltage loss of 2.2 Volts. Now put in a current of zero, and what do you get? Voltage loss of Zero. And it has meaning, with no current flowing the voltage loss is zero, we have the full supply voltage at the far end of wire. Note: No division by zero was done here, no electrons were harmed either.

[trader_4]

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.

[FromTheRafters]

trader_4 brought next idea :

D=RT is a relationship and can be written as T=D/R or R=D/T and it is still the same relationship.

Okay, so if I is zero, what is R? Can you show that the relationship still holds?

Well duh! The thing is that it is *not* "voltage drop" because an open fuse does not dissipate energy.

"Ohm's law states that the *current through a conductor* . . ."

Where's the current through an open fuse, brainiac?

"Voltage drop describes how the supplied energy of a voltage source is reduced *as electric current moves through the passive elements* . . ."

Show me how Ohm's law holds when the current is zero, brainiac.

[FromTheRafters]

Of course there isn't, because there is no current. You can't have a voltage drop when there is no current. Thanks for finally agreeing with me.

[FromTheRafters]

Sorry to have to tell you this, but there can be no power without any current either.

[FromTheRafters]

Yes, it sure does. I always heard of that as power *loss* or copper

*loss* though. Power Drop was a sort of tap which went from the pole to the customer.

The term 'voltage drop' was only used in circuits with current flowing through them. All of the deflections aside, I'm still not ready to believe that a blown fuse has a voltage drop across it no matter what these brainiacs say.

[FromTheRafters]

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.

[FromTheRafters]

They're the same? Tell these guys then, 'cause they have it all wrong.

[Diesel]

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.

[Diesel]

*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. :)

[Diesel]

No, it doesn't. It's about 1.5amps at 24 volts. 35.1 watts. My apologies for the confusion on my end!

used:

ohms=voltage/amperage

240/14.6=16.4 (16.438)

watts=volts(x2)/ohms

24x24/16.4=35.1 (35.121)

amps=watts/volts

35.1/24=1.5 (1.4625)

[Diesel]

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. :)

[clare]

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.

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.

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

[clare]

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.