Another current question

Did you notice the lesser current on L3?
Yeah...I don't know where that comes from. There are many forms of 3 phase. Some with "wild legs" and no neutral... some with a center tap on a 240V leg to provide 120V... I also know that there were a couple of weeks of homework, back in 1989 taking Circuits 2... revolving around balancing deltas and wyes... and that many of them run just a leeeeeeeetle imbalanced, in practice.

I also didn't want to bring up the second table... as that's for a different model, rated for 16kW... I was trying to keep variables under discussion to a minimum (for once)... I'm trying to learn to be a little more sparse of words...lol
 
Wow this is getting entertaining.

I agree with someone above that it was wrong when some people believed just because the nameplate said 15KW or whatever that it must still provide 15KW if voltage was reduced. WRONG

The 15 KW is the power equaled to volts x amps when 240 was applied
However if only 208 were applied it’s NOT going to still produce the same power, current will be less due to ohms law and power will be less no longer 15KW

I know elec topics draw out the most opinions and answers right or wrong but this one is really something
I’m sticking with the answer to the question about a resistive heater (no motor no power questions) has less volts applied the current will be less !!!!! And I’m convinced I’m right as well as many others.

God bless everyone here regardless of their answer to what I think is a simple question about a heater and lower volts

John T
 
Wow this is getting entertaining.

I agree with someone above that it was wrong when some people believed just because the nameplate said 15KW or whatever that it must still provide 15KW if voltage was reduced. WRONG

The 15 KW is the power equaled to volts x amps when 240 was applied
However if only 208 were applied it’s NOT going to still produce the same power, current will be less due to ohms law and power will be less no longer 15KW

I know elec topics draw out the most opinions and answers right or wrong but this one is really something
I’m sticking with the answer to the question about a resistive heater (no motor no power questions) has less volts applied the current will be less !!!!! And I’m convinced I’m right as well as many others.

God bless everyone here regardless of their answer to what I think is a simple question about a heater and lower volts

John T
John, it is mostly folks jawing with each other, as you will notice John908 was last here at post 13 out of the total of 60+
 
John, it is mostly folks jawing with each other, as you will notice John908 was last here at post 13 out of the total of 60+
X2 Jessie. And I thought it was us lawyers who were so wordy lol. Hey it happens anytime anything electrical is posted and I’m as guilty as the next fellow but this one takes the cake and it’s hard to believe anyone would think current increased when voltage decreased given the terms of the question ????? but not everyone has extensive electrical training or is an electrician or engineer with all due respect to any of the fine lay gents here. We can’t all be plumbers or teachers or mechanics or engineers or farmers or carpenters etc etc snd no shame in that. There are several professional electricians and engineers here I appreciate and have learned from them yayyyyy

Nice chatting with you Jessie snd God bless everyone here regardless of their profession and I’m proud to be here and help when I can but my training is only legal and electrical maybe some farming

John T. BSEE, JD
 
nobody actually bothered with performing the math
I guess post #49 doesn't count. And, the OP's question has been answered by a number of individuals - current will decrease running on 208v vs 240V.

Still interesting thoughts from everyone! Keep the thread alive; we have only about 30 pages to go to equal a discussion about an oil-fired furnace on a different forum!
 
Wow. I'm still wondering what the OP, who is retired, is working on. In a thread last month he says he has supply voltage of 240v 3 phase for a kettle and now he has supply voltage of 208v 3 phase for a skillet.

I detect a little (or a lot of) pot stirring here.

I wish he would explain to me which cooks bacon faster without burning it to a crisp, the kettle or the skillet? Either way I suppose they are both black.
 
Wow. I'm still wondering what the OP, who is retired, is working on. In a thread last month he says he has supply voltage of 240v 3 phase for a kettle and now he has supply voltage of 208v 3 phase for a skillet.

I detect a little (or a lot of) pot stirring here.

I wish he would explain to me which cooks bacon faster without burning it to a crisp, the kettle or the skillet? Either way I suppose they are both black.
Guess his commercial appliance jockeying business is taking off.

Wonder if he's using his 700lb Hobart mixer to stir the pot...
 
X2 Jessie. And I thought it was us lawyers who were so wordy lol. Hey it happens anytime anything electrical is posted and I’m as guilty as the next fellow but this one takes the cake and it’s hard to believe anyone would think current increased when voltage decreased given the terms of the question ????? but not everyone has extensive electrical training or is an electrician or engineer with all due respect to any of the fine lay gents here. We can’t all be plumbers or teachers or mechanics or engineers or farmers or carpenters etc etc snd no shame in that. There are several professional electricians and engineers here I appreciate and have learned from them yayyyyy

Nice chatting with you Jessie snd God bless everyone here regardless of their profession and I’m proud to be here and help when I can but my training is only legal and electrical maybe some farming

John T. BSEE, JD
John; Put a volt and ampmeter on a 12 volt battery in a 2 cylinder JD & when it comes up on compression you will find as the voltage drops the amps go up!
 
John; Put a volt and ampmeter on a 12 volt battery in a 2 cylinder JD & when it comes up on compression you will find as the voltage drops the amps go up!
Mornin Teddy. Yep as I understand it when the points break open the coil voltage as seen across the plug gap Ramps up (not 10000 volts suddenly) just high enough until current arc jumps across the plug gap and of course voltage drops off and stored coil energy is extinguished, then start all over pump current through the coil as points are closed then start the above all over !!!!

As you know that’s not the same thing we’re talking of here that given a fixed resistive load if voltsge goes down also will current. DUH I= V/R

Always nice sparky chatting with you have a good day

John T
 
Funny! That example is about as relevant to the question at hand as me saying that my solar array puts out more power in sunlight than in the dark of night.
 
Mornin Teddy. Yep as I understand it when the points break open the coil voltage as seen across the plug gap Ramps up (not 10000 volts suddenly) just high enough until current arc jumps across the plug gap and of course voltage drops off and stored coil energy is extinguished, then start all over pump current through the coil as points are closed then start the above all over !!!!

As you know that’s not the same thing we’re talking of here that given a fixed resistive load if voltsge goes down also will current. DUH I= V/R

Always nice sparky chatting with you have a good day

John T
Not testing the ign curcuit but the starting curcuit John.
 
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
I know enough about 3-phase to leave it to you sparky guys.
A friend sent me this:

The catch is in the fact that 3 phase loads are connected differently. There is a wye connection such as:
image.png

Or the delta connection.
image.png
 
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.
The issue is not total power of the system (what the meter outside reads), but power of one specific device, and for that device, it is simply multiplying I*V.
 

We sell tractor parts! We have the parts you need to repair your tractor - the right parts. Our low prices and years of research make us your best choice when you need parts. Shop Online Today.

Back
Top