No, there won't. The wire isn't a super conductor. It takes a little to push the electrons on it. The thinner and longer the length, the more is lost in transit. If it were a super conductor, what you've written would be absolutely true. As long as the wire has resistance of it's own, we're subject to voltage drop. We have various ways in which to minimize the voltage drop, though. Short of using a super conductor however, we can't outright prevent it.
Not exactly. You can have high voltage and next to no current behind it (think of a taser) or low voltage and a considerable amount of current behind it (think of your car battery) it's no chump and your starter motor isn't exactly a low drain device...You can also lose a few volts in our example, and, still be able to pull enough current to make up the difference.
The easiest way I know of to explain the relationship is this:
It may be useful to consider the image of water in a hose. Voltage is equivalent to pressure, water flow is equivalent to current and the diameter of the pipe is equivalent ot the thickness of the wire - or resistance.
You are correct. If there is no current flowing, there is no voltage drop. The same is true for water pipes. If no water is flowing, the pressure is the same at both ends.
The heating element looks like a resistor. There is no magical making up. Take a simple circuit with a voltage source V and a resistor of value R1. The current is V/R. Now add an additional resistor in series, R2 that represents the resistance of the wire in our system. The current is now V/(R1+R2), which is less than the current of V/R1.
Like Gfre said, if you look at a heating element, if it's rated for two voltages, the power output is stated lower at the lower voltage.
IDK what parameters they used. I worked out the actual numbers, it's very simple and the voltage drop due to the resistance in the wire is only 2.7%, at 20A, for 100 ft of #12. Why no reply to the actual math? Here it is again:
The resistance of 12 gauge wire is .0016 ohms per foot.
100 ft, you have .16 ohms. At 20A, that produces a voltage drop of 3.2V . There are two conductors so double it, 6.4V. It's a 240V circuit, 6.4V drop from 240V is just 2.7%, not 5.6%.
No idea what you're even doing here, no loads? yet 1500?
Again, the resistance of #12 is .0016 per ft. The math has been presented above, the voltage drop at the full 20A is just is just
6.4V, which at 240V is just 2.7%. Even using your higher number, which is probably coming from using some high temperature, you still get a voltage drop of only 3.2%. There is no 125 ft, he said the run is under
100 ft.
And you say this is with no load present yet. That's incorrect too. This is with the MAX load the circuit is rated for, the full 20A. With no load, there is no voltage drop period.
Then use #8, #6, etc. Same argument can be made there. It does nothing in terms of safety or proper operation of the heater.
The closer I can get to their expected input
You're really confused here. The heater does not magically adjust to draw more amps. Take this to an extreme. If I put just 24V across those heaters, following your logic instead of drawing 14A at 240V they would draw
10 x 14A = 140A? It doesn't work that way.
BS. If that was the case, no electrical inspector would approve it. You have one, Gfre, telling you that it does meet code and citing NEC to back it up.
More BS.
at near full load over a period of time is that the connection
And even using those numbers, the voltage drop at the full 20A circuit rating is just 3.2%. At the actual load, 14.6 amps, it's just 2.3%, so again there is no problem, no hot wires.
Show us where Ugly's says it's a violation of NEC, unsafe, etc to wire the circuit on a #12.
I haven't seen any 240volt baseboard heater wired with a 12/2
That depends entirely on what the rated capacity of the heaters is. And if you're that familiar with heating elements you should know that there ones that are rated for either 240V or 120V and that at the lower voltage they put out LESS power, they don't magically adjust, pull more current, to put out the same power they do at 240V.
I think the last time I actually observed that
Which again implies that this is destined to fail, when of course it's not.
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.
Good grief, you need to take electricity 101. Gfre is correct, with no load there is no voltage drop, the full 240V is available and can be measured at the end of the 100 ft of wire. The only way you get voltage drop is with a load and current flowing. Let RL be the load connected to a voltage source V using wires that have a resistance RW. With RL not present the voltage at the open wires where RL would be is equal to V. With the load RL inserted, the voltage across RL is V- RW*V/(RL+RW). You should know this just from common experience. Measure the voltage at a receptacle on a long circuit with no loads on it, you get the same voltage as you do at the panel. Turn all the appliances, lights, etc on and you get some number of volts less, due to the voltage drop.
Resistor is a resistor man. And a heating element looks more like a resistor that an intelligent widget that draws more amps. Here's a though experiment. I take a 3500W heating element that draws 14.6 A at
240V. Now I hook it up instead to a 24V source. Does it draw 146 A?
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.
BS. There is no "transit", the current flow is zero, the voltage across the wires is the same at both ends. The voltage drop for each wire is V = I * RW, where I is the current, RW is the resistance of the wire. Set I to zero, and what do you get? This is a very simple application of Ohms's Law. And if you think this is wrong, explain what the correct formula for the voltage drop is.
Sigh, more FUD.
More wandering in the wilderness. Just apply Ohm's Law to the actual circuit, like I did above.
Yes, it's simple, unfortunately you have it all fouled up. With no load, there is no loss, because no current is flowing, period.
Even more amazing is how people that don't understand Ohm's Law make comments like that.
That I can agree with. There is no "voltage drop" because there is no circuit in which current can flow. Voltage drop is defined in terms of a circuit with current flowing from the source through a load and back to the source and voltage drop is across all of the resistances in the circuit including the internal resistance of the source.
Think of it this way:
Distance equals rate times time (D=RT) and you have two trains on a railroad track. Detroit to Chicago on one end and Chicago to Detroit on the other. Neither train is moving. Does the distance between them drop to zero?
No, it doesn't.
Voltage drop has no meaning in an 'open circuit', which isn't actually a circuit at all, just as 'rate' has no meaning for objects which aren't moving.
From Wikipedia:
"An electronic circuit is composed of individual electronic components, such as resistors, transistors, capacitors, inductors and diodes, connected by conductive wires or traces through which electric current can flow."
If current can't flow, it stands to reason that it is not a circuit, and 'voltage drop' has no meaning.
He's paying attention and his sarcasm is on point.
Except of course heating elements look like resistors and they don't magically readjust to pull more amps with a lower voltage to keep the watts constant. At 240V, a 3500W heating element pulls 14.6A. Following your logic, if we instead put 24V on it, it will pull 146 A and continue to put out 3500W. That makes sense to you?
Give us the formula for that voltage drop in the wire. The rest of us know it's V = I*RW, where I is the current, RW the resistance of the wire. Ergo, I = 0, V = 0. Capiche? That means that with no load, the full
240V is present at both ends of the long wire. Now put even a tiny load on and now you have voltage drop. The larger the load, the larger the voltage drop in the wires, because the current is higher.
Again obvious you don't understand the basics. Volts are not "lost" If they are and you're so smart, give us the formula for the "lost" volts with no current flowing.
Might I suggest a course in electricity 101? Understand how to apply] Ohm's Law to a simple circuit? Kirchoff's Laws?
It does have meaning, it's ZERO. Zero in physics has meaning. It can be measured. The voltage on the ends of that 100 ft of wire is exactly the same., because the voltage drop across the wire with no current flowing is zero. It's like saying a velocity of zero has no meaning.
There is no such thing as a 'voltage drop' when there is no current flowing.
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D=RT, so when the rate (velocity) is zero then D=zero*T and since zero*T equals zero then D equals zero, so you are saying that the distance is zero. This is obviously not the case.
Consider this:
What is the amperage (A) when you put zero ohms (R) resistance across an ideal 12 volt (V) source?
Answer:
Undefined! Because A=V/R and you can't divide by zero.
This is simple math, why are you struggling with it?
The distance in your formula is not the distance between the cities, which of course is constant and never changes. The distance in your formula is the distance THE TRAIN TRAVELS.
Which is exactly the same thing as saying the voltage drop is zero. It's like saying if we have 3 apples and take 3 away, how many are there now? The answer of course is ZERO, but what you're arguing is that ZERO has no meaning. It does with apples and it does with regard to Ohm's Law and a voltage drop of zero with no current.
Say what now? Distance Traveled= Rate x Time. If the rate is zero, then the distance traveled is zero. Again, clearly zero has meaning.
Poor attempt at diversion. But we're not dividing by zero in the case of Ohm's Law, nor in your example of motion above. And from a math, physics and engineering perspective what happens in your new example is that as the resistance approaches zero, the current approaches infinity. We deal with infinity and things approaching limits in engineering and math. It's not a mystery. But none of this has anything to do with what we are doing with Ohm's Law, because the voltage drop does not involve dividing by zero.
Look, now you're going to start taking cheap shots? Tthe only one here who was obviously struggling is Diesel. He doesn't understand Ohm's Law. But now we can add you to the list because you can't understand that in the simple case of distance = rate * time, a zero rate gives an answer of ZERO distance traveled and that answer of zero definitely has meaning, just like zero voltage drop from Ohm's Law has meaning.
Gfre, WTF has happened to education in America? We now have two idiots here trying to explain basic math and electricity to us and neither knows WTF they are talking about.
IIRC, it's I squared R. Of course, it's still no voltage drop with no current. Is it possible there's a confused poster here, who has R and thinks it's E.
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