Voltage Up, Current down......but not always?

[11bgrunt]

Utilities will use voltage reduction at the substation transformer regulation to shave peaks.
Load is fixed. As many of you have shown, if you fill in the blanks of the equation and do not change the load, voltage down equals current down.
Easy and cheap.

 

[Besoeker]

Conversely a predominantly inductive-load (eg, motor) decreasing the source voltage will result in an increase of load-current if the driven-load is constant, i.e., essentially independent of relatively small changes in motor speed

Although it's a commonly held view, it is not always the case.
If the motor is running at part load, reducing the voltage can reduce the current. It's easy to see this if you start with the no load condition. All the current goes in the Xm part. Reducing the voltage will certainly reduce that.

 

[iwire]

Guys, am going to point out that the OP has not replied at all yet While at the same time you all have had fun flexing your extreame engineering knowledge to each other.

I really wish we could keep it simple until the OP comes back to ask more questions.

In my personal opinion most of the replies here chase people off who are not so deep in theory.

 

[iwire]

I've always been taught this to be the case, but have recently been told otherwise. Can someone enlighten me as to why a large load bank would drawing less current at 208v than it would at 240v? I'm thinking it has something to do with resistive vs reactive loads, but that's about as much as I know.

Thanks for the help. :?

Larry,

A load bank is just an electric heater, if you reduce the voltage to it the current will drop ....but the heat output will also decrease.

If you had to get the same heat output from a lower voltage the heating element would have to be changed and it would draw higher current at the lower voltage to make the same heat output.

Does that help?

Feel free to ask questions.

 

[GoldDigger]

Larry,

A load bank is just an electric heater, if you reduce the voltage to it the current will drop ....but the heat output will also decrease.

If you had to get the same heat output from a lower voltage the heating element would have to be changed and it would draw higher current at the lower voltage to make the same heat output.

Does that help?

Feel free to ask questions.

One possible source of confusion is that some specialized load banks are not just resistances.

For example when measuring the capacity of a battery (such as the 20 hour discharge rate, C₂₀), you need to keep drawing a constant load current as the battery voltage decreases. So for that you need a regulated constant current load.
For other specialized purposes you want a constant power load instead, so you incorporate a regulator which increases the current as the input voltage drops.

But for common electrical testing, all that is needed is a way to waste a (large) approximately known amount of power safely and so a simple resistive load bank is what is used.

 

[qcroanoke]

Guys, am going to point out that the OP has not replied at all yet While at the same time you all have had fun flexing your extreame engineering knowledge to each other.

I really wish we could keep it simple until the OP comes back to ask more questions.

In my personal opinion most of the replies here chase people off who are not so deep in theory.

Thank you. Iwire.

I enjoy reading some of the replies and the discussions on some threads but sometimes it is just way over my head.

 

[qcroanoke]

Larry,

A load bank is just an electric heater, if you reduce the voltage to it the current will drop ....but the heat output will also decrease.

If you had to get the same heat output from a lower voltage the heating element would have to be changed and it would draw higher current at the lower voltage to make the same heat output.

To Quote Howard Johnson from Blazing Saddles "now who can argue with that!"

 

[iwire]

One possible source of confusion is that some specialized load banks are not just resistances..

Sure, but why not just keep things simple until the OP asks more questions?

I tend to be blunt, so here it is. I think this thread is a great example of members trying impress each other more than actually trying to help the OP.

That was not aimed specifically at GD or anyone, and I am certainly not disputing what has been said, and I really do value the wealth of knowledge offered by those above that have posted.

All I am asking is keep it simple until more questions are asked.

 

[Phil Corso]

Iwire!

Bravo!

Regards, Phil Corso

 

[infinity]

Sure, but why not just keep things simple until the OP asks more questions?

I tend to be blunt, so here it is. I think this thread is a great example of members trying impress each other more than actually trying to help the OP.

That was not aimed specifically at GD or anyone, and I am certainly not disputing what has been said, and I really do value the wealth of knowledge offered by those above that have posted.

All I am asking is keep it simple until more questions are asked.

I agree, keep it simple. Would you be more likely to die if you contacted a 480 volt circuit or a 12 volt circuit? Higher voltage, higher current, higher probability of death.

 

[Interplex]

Voltage up -- current down

Voltage up -- current down

It all boils down to not including all the parameters in the orriginal supposition.
(My first post - a requirement as a new member).

W2NH

 

[qcroanoke]

It all boils down to not including all the parameters in the orriginal supposition.
(My first post - a requirement as a new member).

W2NH

Welcome to the forum!

 

[K8MHZ]

Welcome to the forum!

And 73!

 

[motormuff]

It boils down to ohms law

It boils down to ohms law

I've always been taught this to be the case, but have recently been told otherwise. Can someone enlighten me as to why a large load bank would drawing less current at 208v than it would at 240v? I'm thinking it has something to do with resistive vs reactive loads, but that's about as much as I know.

Thanks for the help. :?

Determining the current draw, on a 240 volt rated water heater that is operated on 208 volts is: (208/240)=.866 Square it, Since power varies with the square of the voltage(50% voltage equals 25% out) we now take (.866 X .866)=.751 If the element is 3000 watts at 240 Volts multiply .751*3000 = 2253 watts if operated on 208 volts. Using Ohms law the same result can be found.