Another current question

Wrong! You might be about right IF he had single phase device, but not even in same ball park for his three phase device. Posters really should not be trying to give advice on things they know nothing about.
S
Again, three phase does not have magical properties that go against the clear and simple proportional relationship between voltage, current, and resistance which is described by Ohm's Law.
 
Again, three phase does not have magical properties that go against the clear and simple proportional relationship between voltage, current, and resistance which is described by Ohm's Law.
Power in 3 phase system is not calculated by simply multiplying 208 or 240 times current. No “magic”, but you do need the correct formula.
 
I was at a campground where some neighbors were trying to cook a big breakfast using two cheap "1500W" Walmart electric griddles plugged into the power post typically used by RVs. They plugged both griddles into the 15A circuit on the power pole, and could not figure out why the griddles were not getting hot. They got warm but they would not get hot enough to cook the bacon and eggs and pancakes.

I keep a Kill-A-Watt meter in my camper which shows Volts, Amps, Watts, etc. of any device plugged into it. When I plugged the Kill-A-Watt meter in between the outlet and one of the griddles, it read 90 Volts and around 9 Amps. Well there's your problem lady. I unplugged the other griddle, and the volts shot up to 120, the Amps shot up to 12, and the bacon started sizzling.

Less Volts. Less Amps. It's that simple.
 
I just arbitrarily picked a resistance value because it is the same in either case just to demonstrate the relationship.

The skillet is only 15KW at a specific voltage, which is no doubt documented on the skillet itself because this is clearly an industrial cooking device.

There is no "magic" involved because it's three phase. It is NOT an inductive load.

I did some research and came up with a cooking device called a "Tilting Skillet" which come as electric or gas powered. In the spec sheet for the electric powered one I found, they were available with heating elements for different power supplies.

https://www.centralrestaurant.com/c...llet-manual-40-gallon-208v-p25j-003-208v.html
Correct! I did a quick research too for electric skillets and found one that ships rated for 240v 3p but can be field converted to 240v 1p, OR they can ship it 208v 3p but field convertible to 208v 1p. In order to keep the wattage rating, the element must be swapped out if using a different voltage.
 
Power in 3 phase system is not calculated by simply multiplying 208 or 240 times current. No “magic”, but you do need the correct formula.
AGAIN, this is for a purely resistive circuit. A heating element is just a plain resistor, and power IS calculated by simply multiplying voltage and current. The phase is not what determines how to calculate, it's the load, whether plain resistive, OR if it has a capacitive or inductive reactance, then you can have shifts in power.

You keep telling us we're wrong, so go ahead and show us what equation you are using. I showed my work, where's yours?
 
Power in 3 phase system is not calculated by simply multiplying 208 or 240 times current. No “magic”, but you do need the correct formula.
We're not talking about power here. We are talking about the current drawn by a pure resistance load, and we're only concerned with less or more NOT the exact value.

It's proportional. The formula for calculating power is irrelevant as long as the same formula is used in both calculations.
 
We're not talking about power here. We are talking about the current drawn by a pure resistance load.

It's proportional. The formula for calculating power is irrelevant as long as the same formula is used in both calculations.
Correct! And that is where these other commenters are getting it wrong, thinking the power rating stamped on the skillet must be maintained regardless what voltage is applied. A 15kw skillet will only get that much power IF it is connected to the 240v supply that it's rated. If connected to a 208v source, it will not pass as much current (and therefore will not reach its 15kw power rating).
 
Correct! And that is where these other commenters are getting it wrong, thinking the power rating stamped on the skillet must be maintained regardless what voltage is applied. A 15kw skillet will only get that much power IF it is connected to the 240v supply that it's rated. If connected to a 208v source, it will not pass as much current (and therefore will not reach its 15kw power rating).
UNLESS, as you stated elsewhere, the element is swapped out for one designed to draw 15KW at 208V.

Then and ONLY then will the skillet draw more current on 208V than 240V.
 
AGAIN, this is for a purely resistive circuit. A heating element is just a plain resistor, and power IS calculated by simply multiplying voltage and current. The phase is not what determines how to calculate, it's the load, whether plain resistive, OR if it has a capacitive or inductive reactance, then you can have shifts in power.

You keep telling us we're wrong, so go ahead and show us what equation you are using. I showed my work, where's yours?
I said draws less current on 208 than on 240 right from the beginning. But first n16ty and then you make statements about the power and calculations such as 15 kW divided by 240v yields 62 amps. That is simply wrong on 3 phase system. Should be using P=1.732 x V ( line to line) x I.
 
I have a generator load bank that is rated for 20KW at 240V. If I have a 120V input, that 20KW load bank at 240 volts is now a 5 KW load bank at 120V. The resistance does not change, only the voltage. Ohms law says that the current will go down with lower voltage at the same RESISTANCE.
 
I said draws less current on 208 than on 240 right from the beginning. But first n16ty and then you make statements about the power and calculations such as 15 kW divided by 240v yields 62 amps. That is simply wrong on 3 phase system. Should be using P=1.732 x V ( line to line) x I.
You said "less" and expect everyone to go by your say-so.

I proved it is less, using Ohm's Law. On the off chance you are addressing me in this statement, I never said anything about power in my proof, because I know power is a different thing. In fact I have repeatedly said that this is not about power.
 
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You said "less" and expect everyone to go by your say-so.

I proved it is less, using Ohm's Law. On the off chance you are addressing me in this statement, I never said anything about power in my proof, because I know power is a different thing. In fact I have repeatedly said that this is not about power.
Not addressing you!
 
My last post was on a kettle and the data plate states 3ph 208 24kw 66.6 amps, so i am wondering how the skillet at 3ph 15kw 240 would draw between 60-70 amps? The skillet does not for some reason show a current spec.
Current can always be calculated if the power and voltage is known. It is typical to list only two out of the three on any piece of equipment, since if you know 2, then you know 3. (power, voltage, amperage)
 
You said "less" and expect everyone to go by your say-so.

I proved it is less, using Ohm's Law. On the off chance you are addressing me in this statement, I never said anything about power in my proof, because I know power is a different thing. In fact I have repeatedly said that this is not about power.
Power is always there whenever there is voltage and current. You can always calculate one of the three if any two are known. The skillet in question has only 2 known published numbers, the 15kw and the 240v. This is about power, but only in the first step, for the purpose of using that number to calculate the resistance of the heating element and/or calculate the current. Knowing current is for determining appropriate wire and breaker size, and knowing power is for how well it can cook food.
 
This time I have a skillet that is 240 3ph 15kw,if I provide a supply voltage of 208 does the current increase or decrease?
Current can always be calculated if the power and voltage is known. It is typical to list only two out of the three on any piece of equipment, since if you know 2, then you know 3. (power, voltage, amperage)


Guys...

A couple things:

3 Phase stuff...
Power is rated as the total from all three phases (15Kw, in this case).
Current is rated "per phase"
Voltage is rated "phase to phase"

To calculate an exact value of per phase current... you need to first divide the total power by 3. Then... your next step, needs to know if the phases are connected as a delta or wye. Wye? Because, in a Wye configuration, the per phase power is per phase current multiplied by the phase to neutral voltage for each phase... phase to neutral? Nobody talked about phase to neutral...darnit!

If you don't know that (delta or wye), you can't go from a power rating to an exact current rating per phase.

But anyway... that's why the projected current draws here all seem so high. There was no division by three first to get the per phase power. (or... in some cases, like JMOR's... the square root of three is used with some assumptions to take care of this...which is legit...but even this makes an assumption of delta or wye)

Generally, if you went from a power source that was 240V phase to phase (for all three phases)... to a power source that is 208V phase to phase (for all three phases)... and connected the SAME load in the SAME configuration (delta or wye), as connected to the 240V ... the per phase current should go down (be less).

I wouldn't, myself, say any more about the exact value of what that current is, since there are too many pieces of information missing and I'm not familiar myself with 3 phase systems with 240V phase to phase for all three phases. As a matter of fact, I don't bet, but I would put equal odds on this being a straight up 240V device...as opposed to a "240 3ph"... I wouldn't believe this without seeing a wiring diagram or some other information.
 
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My answer. Decrease

OMG I just got here and surprised to so many responses to what I was taught in science and physics and electrical engineering as a simple matter of fact

Ohms law and the question was about a heater which is a pure resistive load

Far as I know Current I = V/R.
Lower volts yields lower current
Say r was one ohm and v was 10 volts then I = 10/1 = 10 amps.
Say v was 100 volts then I = 100 amps
Again less volts means less current

The question didnt mention motors or power so no need to go there

Goes to prove what I always say. Any electrical question draws many including me out of the woodwork lol Everybody has an answer of some sort but often it’s not an answer to the question asked. Don’t get me wrong there’s a fine bunch of gents here all trying to help which I enjoy appreciate and try my best to help. The question involved a heater resistive load and didn’t ask about power or motors only if voltage was less would current be less I’m retired and could be wrong on
My answer but believe it’s still true current is less if voltage is less

John T. BSEE but retired so no warranty
 
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