20 amps at 110 volts Bandsaw - Page 4 - Woodworking Talk - Woodworkers Forum
Old 04-16-2016, 03:21 AM
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Originally Posted by Tony B View Post
Not quite, if we are talking the SAME circuit.
If a tool pulls 20A @120V, the same tool will pull only 10A at 220V.
The 20A running through a 12Ga wire will be warmer than 10A running through the same 12 ga wire.
With the higher current in the 120V circuit, we will also have a higher temp resulting in a higher resistance which will turn into a higher voltage drop reducing the efficiency of the 120V circuit. This gets compounded by the motor surge current also.
If an electrician were to wire a new 220V circuit he will most likely use at least a 10 ga wire and further decrease the line voltage drop resulting in better starting and running.
Your electric bill will not reflect these factors because they are usually minute changes and for short durations.

Bottom line is that if you used the same wires in your wall and changed the voltage from 120 to 240, you will be drawing only half the current.
Absolutely correct. but go back and read the post I was correcting and what I wrote. The point I'm trying to relay is that 20 amps at 120, or 20 amps at 240 will produce the same heat in the same size conductor.
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Old 04-16-2016, 06:15 AM
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Originally Posted by ryan50hrl View Post
Like for example, that 14 gauge wiring will handle 15 amps just fine....Attachment 232626
This is where you and I will always disagree. Too many electricians are using 14 gauge wire for outlets. Yesterday I had to repair a fence using a smudgepot air compressor. I tried plugging it into the outside outlet by their front door and there was barely enough power there to make it try to start. I ended up having to carry the compressor around the house to the garage every time it needed to cycle. I'm sure someone thought, it's just an outside outlet all they will be using on it is Christmas lights.
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Old 04-16-2016, 06:27 AM
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Originally Posted by Steve Neul View Post
......... I'm sure someone thought, it's just an outside outlet all they will be using on it is Christmas lights.
Yes, and how much did they save off the total cost of the house by using lighter wire?

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Old 04-16-2016, 06:34 AM
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Originally Posted by d_slat View Post
......The point I'm trying to relay is that 20 amps at 120, or 20 amps at 240 will produce the same heat in the same size conductor.
Sorry, yes you are correct. I took it out of context because previously we were comparing the current from one particular motor on 120 vs. 240. My bad.

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Old 04-16-2016, 07:46 AM
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14AWG will handle 15 amps, yes. No problem. Residential "industry standard" is to run 14AWG for lighting circuits and 12 AWG for receptacle circuits.

A motor that draws 15 amps should be fed by a 20 amp circuit, or it will trip a breaker when the motor is under full load.

Last edited by BZawat; 04-16-2016 at 07:49 AM.
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Old 04-16-2016, 09:54 AM
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Originally Posted by Steve Neul View Post
This is where you and I will always disagree. Too many electricians are using 14 gauge wire for outlets. Yesterday I had to repair a fence using a smudgepot air compressor. I tried plugging it into the outside outlet by their front door and there was barely enough power there to make it try to start. I ended up having to carry the compressor around the house to the garage every time it needed to cycle. I'm sure someone thought, it's just an outside outlet all they will be using on it is Christmas lights.

I will be the first one in this thread to admit they do NOT know what they are talking about, I am just a lowly carpenter.

Can someone explain the following to me- We often run 50-100ft of 12g extension cord to where we are working on a job. We often have both my compressor and either my CS or MS both plugged in to that one cord. It's not uncommon to have the compressor cycling while someone is making a cut. My compressor is rated at 12A and both my CS and MS are rated at 15A. Why have I never tripped a breaker on a single 15 or 20A recepticle regardless of 14 or 12g wire used in the circuit?
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Old 04-16-2016, 09:58 AM
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Think of it this way. A new truck today has somewhere around 350 horsepower. But you rarely use 100% of that power. Your most often probably only using 150 hp as you cruise down the highway.

Your saw and compressor are the same, accelerating both the saw and compressor motors from a dead stop to running at the same time likely would trip the breaker, but you rarely if ever do that.

Also cutting 2x4 pine, or trim boards isn't very taxing on your saw. If you were trying to cut 4x8 hard maple, your saw would be working a lot harder.

The tools don't make the craftsman....
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Old 04-16-2016, 10:17 AM
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Originally Posted by ryan50hrl View Post
Think of it this way. A new truck today has somewhere around 350 horsepower. But you rarely use 100% of that power. Your most often probably only using 150 hp as you cruise down the highway.

Your saw and compressor are the same, accelerating both the saw and compressor motors from a dead stop to running at the same time likely would trip the breaker, but you rarely if ever do that.

Also cutting 2x4 pine, or trim boards isn't very taxing on your saw. If you were trying to cut 4x8 hard maple, your saw would be working a lot harder.

Actually initial start ups at the same time happen more often than you might think. Probably not enough to say 'regularly', but I've seen it happen more times than I can count.

That aside, if the compressor is cycling, using IDK lets say half of the amperage it's rated for at initial start up, and I pull the trigger on my MS, I would think the initial 15A momentary draw from the saw along with the steady 6A'ish draw from the compressor would be enough to trip a 15-20A breaker. Or are all circuits made to handle momentary draws well beyond their ratings?
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Old 04-16-2016, 10:24 AM
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Your counting on the initial start up as a second, or the moment you hear it. The initial startup draw happens so quickly your unable to register that time frame without equipment to monitor in fractions of a second.

The tools don't make the craftsman....
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Old 04-16-2016, 12:52 PM
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Originally Posted by Chamfer View Post
I will be the first one in this thread to admit they do NOT know what they are talking about, I am just a lowly carpenter.

Can someone explain the following to me- We often run 50-100ft of 12g extension cord to where we are working on a job. We often have both my compressor and either my CS or MS both plugged in to that one cord. It's not uncommon to have the compressor cycling while someone is making a cut. My compressor is rated at 12A and both my CS and MS are rated at 15A. Why have I never tripped a breaker on a single 15 or 20A recepticle regardless of 14 or 12g wire used in the circuit?
Think of electricity like it was water. If you were trying to use a 1/4" air hose for a garden hose you would have a lot of trouble getting the water you need. Then the farther you go with that undersized hose the worse the problem would be. It's not so much tripping the breaker but having enough power to do what you need. The electrons in electricity pass over the outside of the wire and the larger surface it has the power it will deliver. Then stranded wire conducts better power than solid copper because the electrons pass over each strand so it has more surface area. Apparently where you have been working you have been plugging in your extension cord into a good source of power. Your 12g stranded extension cord is also a lot better conductor than a 14g solid wire used in romex wire.
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Old 04-16-2016, 04:39 PM
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Had to try out my kilo-watt meter on my delta contractor table saw 1.5 hp
Dedicated 20 amp 12 gauge circuit.
Heavy duty extension cord .
121 volts at the meter (at the end of the extension cord).
The meter flashed around 22 amps then 23 amps in quick succession and then 4.49 amp as it came to speed.
Cutting a 1/2 piece of plywood the amps increased to around 6amps.
So even with capacitor start you still get a spike but a very short one and not enough to trip the breaker.
Light cuts and you will not draw the maximum rated amps of the unit.
Just because a breaker doesn't trip means your using safe practices.
I have a couple of circuits in my basement I've derated at the breaker ( #12 wire with 15 am breaker)..why ?
So someone long after I'm gone doesn't overload the circuit..say like a continuous draw of a 15 amp space heater
and god only knows what else they can think of ( treadmills , tvs ect).

The National Electrical Code doesn't limit the number of receptacles you can place on a 20-amp circuit, but you'll overload the breaker if you run appliances that draw more current than the breaker can handle. The NEC does specify that a circuit breaker shouldn't handle more than 80 percent of the load for which it is rated unless the breaker is labeled otherwise. By this standard, the total current draw on a 20-amp circuit shouldn't exceed 16 amps. This allows the breaker to handle the temporary surge that happens when an appliance such as a power saw or air conditioner starts up.

Knot Stumped ...just confused once in a while.
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Old 04-16-2016, 04:56 PM
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One reason that a breaker may not trip when using long extension cords is that the long wires act as series resistance and lower the voltage delivered to the load. As to breakers, they operate on a magnetic/time constant curve. Any current in excess of the rated current will "eventually" trip a breaker. A small overload may take many minutes to trip, and as the overload increases, the time to trip shortens. This is the reason the 20 amp circuit can supply very short duration currents of 25 or 30 amps or more.

If you look at the charts tool manufacturers supply with most tools, the current and the distance both play a role in wire size. For example a tool that will run on a #14 extension cord at 25 feet or 50 feet may require a jump to a #12 wire at 100 feet to insure adequate voltage to the tool. Another thing often overlooked is the distance from panel to outlet. In a big shop this can be 40 or 50 feet or more, and adding an extension cord just further acerbates the voltage drop.

Personally I don't use #14 cords for any power tool, #12 may be overkill for a 1/4: drill, but not having a lighter cord around does tend to prevent grabbing one when using a larger tool.

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Old 04-17-2016, 04:34 AM
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Originally Posted by d_slat View Post
Ohms law does not apply to the heating of a conductor like you are trying to describe here, since you won't see 120 or 240 volt drop across the conductor. Current is what heats the conductor, so 20a at 120v or 240v will produce the same heat in the conductor.

As I said in my post you are referring to, heating rate for a given load (resistance) is proportional to the square of the current times the resistance:

Heating rate = current x current x resistance

But Ohm's law relates the resistance as the ratio of voltage and current:

voltage = current x resistance

So the heating rate (by the virtue of Ohm's law) becomes the product of the voltage and the current:

Heating rate = current x voltage

which is what we understand by the 'power' flowing through the circuit.

If you read carefully, you will find both of us are saying exactly the same thing.

BTW, the reason an overload beyond 20A may not be tripping the breaker is because a factor of safety (1.5) is built-in. Thus I think what is labeled as a 20A breaker, is actually a 30A breaker.

Keep thy axe sharp.

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Old 04-17-2016, 07:11 AM
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If "google" stranded wire vs solid core wire you will find out that at 60hrts there is virtually no difference between stranded and solid core wire regarding resistance. The skin effect comes into play at much higher frequencies.

The main reason for using stranded wire is that in larger sizer it is more flexible.

George
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Old 04-17-2016, 08:10 AM
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Originally Posted by Jig_saw View Post
If you read carefully, you will find both of us are saying exactly the same thing.
No, we're not. Your inclusion of ohms law into what we are talking about is dead wrong. Voltage has no significant affect on the heating of a conductor.

Edit: I think I see what you are trying to do, but it doesn't work the way you are thinking. If you want to substitute voltage into the equation for heating of the conductor instead of current x resistance, you have to use the voltage drop across the conductor at load, not the circuit voltage.

Last edited by d_slat; 04-17-2016 at 08:26 AM.
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Old 04-17-2016, 10:03 AM
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Originally Posted by Jig_saw View Post
As I said in my post you are referring to, heating rate for a given load (resistance) is proportional to the square of the current times the resistance:

Heating rate = current x current x resistance

But Ohm's law relates the resistance as the ratio of voltage and current:

voltage = current x resistance

So the heating rate (by the virtue of Ohm's law) becomes the product of the voltage and the current:

Heating rate = current x voltage

which is what we understand by the 'power' flowing through the circuit.

If you read carefully, you will find both of us are saying exactly the same thing.

BTW, the reason an overload beyond 20A may not be tripping the breaker is because a factor of safety (1.5) is built-in. Thus I think what is labeled as a 20A breaker, is actually a 30A breaker.
No, it's actually a 20 amp breaker. See chart below for how breakers work. A 30 amp breaker will have higher trip currents for the same time factor than a 20 amp. It is not a "safety factor".
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Originally Posted by GeorgeC View Post
If "google" stranded wire vs solid core wire you will find out that at 60hrts there is virtually no difference between stranded and solid core wire regarding resistance. The skin effect comes into play at much higher frequencies.

The main reason for using stranded wire is that in larger sizer it is more flexible.

George
Yes, and it's hard to convince people of this. They read it somewhere, and didn't notice or dont remember the part frequency plays in skin effect.

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Originally Posted by d_slat View Post
No, we're not. Your inclusion of ohms law into what we are talking about is dead wrong. Voltage has no significant affect on the heating of a conductor.

Edit: I think I see what you are trying to do, but it doesn't work the way you are thinking. If you want to substitute voltage into the equation for heating of the conductor instead of current x resistance, you have to use the voltage drop across the conductor at load, not the circuit voltage.
True, you can't use current at applied voltage to figure heat loss. Most of the power consumed is at the load, only a small portion is lost in heating the conductors. The only time when the heat in the conductor becomes a factor is when the wire is too small for the load current, and the heat is enough to damage the conductor.

When a conductor becomes appreciably warm to the touch, not only is it too small, but it is causing more heat buildup in any motor it's supplying. Not a factor in resistive loads, just a loss of efficiency.
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