Another current question

[JOHN908]

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?

 

[38 Sears Roebuck]

Wow, if you have an electric skillet that's 240 volts/3 phase? I bet that cooks food in no time?

 

[504]

I had to rewire the heating elements on the dishwasher to get it to work on 220v. I would expect your pan to heat slower, but I would expect it to work fine. I am not nor ever been an electrician . Go Chiefs

 

[JMOR]

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?

Decrease

 

[m16ty]

The current will go up on 208v vs 240v.
Any time you apply a lower voltage to the same load, the current will always go up (Ohm’s Law).

 

[m16ty]

Decrease

No, current will go up on 208v vs 240v.

 

[BarnyardEngineering]

The current will go up on 208v vs 240v.
Any time you apply a lower voltage to the same load, the current will always go up (Ohm’s Law).

If you're citing Ohm's Law, you have it backwards. Ohm's Law is for simple resistive loads, like an electric skillet.

V=IR

Solving for I, the current

I = V/R

R is the resistance of the electric skillet. That does not change, so any value can be used. Let's say the resistance is 8 ohms.
V is voltage, 240 vs. 208

I = 240 V/8 Ohms = 30 Amps
I = 208 V/8 Ohms = 26 Amps

For an inductive load you would be correct, but then it has nothing to do with Ohm's Law.

 

[JMOR]

This is a forum, so you can get both or all, information, misinformation, opinion, BS, etc.

 

[da.bees]

62 amps at 240v, 72 amps at 208v.

 

[wore out]

62 amps at 240v, 72 amps at 208v.

LOL!

 

[joeysmith]

62 amps at 240v, 72 amps at 208v.

Hell of a skillet

 

[m16ty]

If you're citing Ohm's Law, you have it backwards. Ohm's Law is for simple resistive loads, like an electric skillet.

V=IR

Solving for I, the current

I = V/R

R is the resistance of the electric skillet. That does not change, so any value can be used. Let's say the resistance is 8 ohms.
V is voltage, 240 vs. 208

I = 240 V/8 Ohms = 30 Amps
I = 208 V/8 Ohms = 26 Amps

For an inductive load you would be correct, but then it has nothing to do with Ohm's Law.

Ohm’s Law also covers current.
Also, your amps are wrong. At 15kw the skillet will draw 62.5 amps at 240v and 72 amps on 208v. Watts divided by volts will get you your amp draw. Ohms law covers this.

 

[JOHN908]

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.

 

[JMOR]

Reading some of the replies since I last posted has me thinking, I need to re-post!
"This is a forum, so you can get both or all, information, misinformation, opinion, BS, etc."

 

[Texasmark]

The current will go up on 208v vs 240v.
Any time you apply a lower voltage to the same load, the current will always go up (Ohm’s Law).

Say what?

 

[DanielW]

Am I interpreting this wrong? I think there's some confusion between the Power Law and Ohm's Law, and also what variables will stay constant.

Power Law: Power = Current X Voltage

Ohm's Law: Voltage = Current X Resistance

You can solve for any variable and mix/match as you please. Ex., Rearrange Ohm's law as: Current = Voltage/Resistance. And sub that eq'n for current into the power law: Power = (Voltage * Voltage)/Resistance ( or P = V^2 / R - the formula for power dissipated by a resistor).

I believe where some folks are running into trouble is assuming what variables are going to be constant. It's fairly safe to assume the skillet is a basic resistive element. And obviously the resistance isn't going to change (because the element is not changing). So current decreases as per Barnyard's and JMOR's responses above.

I think where m16ty ran into trouble was assuming the 15 kW stayed constant. That's not the case. Not for a resistive load, anyway. The 15 kW would be the skillet's nameplate power rating at the nameplate 240 Voltage. But decrease the voltage, and the power will also decrease. And as per the formula derived above, it will decrease by a factor of 208^2 / 240^2 (thus: new power at 208V will be about 75% of the nameplate 240V power - about 11.3 kW)

If it's three-phase and AC, there are other factors to consider (power factor, effect of phase angles, etc.) so that power formula I mentioned above doesn't strictly apply verbatim. And if we had a constant power or inductive load then the current would indeed increase to maintain the same power. But that's the basic idea for a basic resistive load, where power's not going to be maintained constant. And regardless of phase effects: Current (and power) will still decrease in three phase - just not quite as per the formulae above.

But it's been a long time since I had to think about these things, so someone tell me if/how I'm wrong.

 

[TheBartman47]

Ohm’s Law also covers current.
Also, your amps are wrong. At 15kw the skillet will draw 62.5 amps at 240v and 72 amps on 208v. Watts divided by volts will get you your amp draw. Ohms law covers this.

If you reduce voltage, amperage will also go down, (and likewise power goes down). An electric skillet is purely resistive load. If it is rated 15kw with a 240v supply, then that's 62.5 amps of current. Ohms law written for resistance is R=V/I. This means the skillet has a resistance of 3.84 ohms. If the only change is dropping the voltage down to 208, then ohms law in terms of current is I=V/R which means the skillet will then be passing 54.17 amps, which then means it's only outputting 11.27kw of power (P=I*V).

 

[JMOR]

If you reduce voltage, amperage will also go down, (and likewise power goes down). An electric skillet is purely resistive load. If it is rated 15kw with a 240v supply, then that's 62.5 amps of current. Ohms law written for resistance is R=V/I. This means the skillet has a resistance of 3.84 ohms. If the only change is dropping the voltage down to 208, then ohms law in terms of current is I=V/R which means the skillet will then be passing 54.17 amps, which then means it's only outputting 11.27kw of power (P=I*V).

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

 

[TheBartman47]

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

Nope, I am correct. My degree is Electronics Engineering Technology. If the only thing changing is the voltage decreasing, then current MUST also decrease. 240v 3-phase is very rare, but might be possible that's what he has. Some erroneously call 240v single phase "three phase" when they're using two legs of 120 to get that 240. If it is true that he actually has 240 3-phase, then connecting it to 208 3-phase is only reducing the voltage to a purely resistive load.

 

[BarnyardEngineering]

Ohm’s Law also covers current.
Also, your amps are wrong. At 15kw the skillet will draw 62.5 amps at 240v and 72 amps on 208v. Watts divided by volts will get you your amp draw. Ohms law covers this.

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