Non-changing current with C-EMF

[Gryd3]

@Arouse1973 but most constant current sources will apply massive power, in this case 10kW to sustain that current.
Maybe using a dc/dc converter could work! Applying 300W input and get an output in that range?

If you account for the resistance in the power source, the actual current induced by the EMF is far lower than we are writing down. This would drastically reduce the power consumption you are concerned with.

 

[XRZ]

@Gryd3, could you use an example(need numbers to visualize!).

 

[XRZ]

Btw, what do all think of my suggestion to use a dc/dc converter? With a 300W input power to sustain 100A @3VDC? Does it make any sense?Could it work?

@Arouse1973 , @Gryd3

 

[Gryd3]

http://powersupply.blogs.keysight.com/2013/06/dc-power-supply-output-impedance.html
Found this. Along with another quick search, it looks like regulated supplies can have internal effective resistances that vary from 1Ω up to 1MΩ depending on the load being driven. This value will change as the load changes...

 

[XRZ]

Making the issue of power negligible! Dropping the 10,000W required input significantly!
No need for 100V x 100A I guess !
Okay, it seems that it is indeed possible to solve such a problem, and not exhaust enormous power. I just wanted a solution where I can add more voltage to a system, without adding x100 or x1000 times of power. Like the example I proposed, where back-emf induced is 2V and the source voltage has to be 3V to maintain 100A, that should require a power source of 300W, but due to the resistance of the circuit the ideal input would be 100A @ 100VDC, but that's a lot more than what we need! So it does make sense to use other methods to reduce power.

 

[Gryd3]

I think we're all on the same page... know there was something simple we were missing.

Thanks @Arouse1973 for filling the blank spot in my head!

 

[XRZ]

But, if you think about it a bit.. how much power do we actually need a factor of what ?
How can we calculate the power that's going to be exhausted additionally to maintain the same current when back-emf is induced by a magnetic field?

 

[Gryd3]

But, if you think about it a bit.. how much power do we actually need a factor of what ?
How can we calculate the power that's going to be exhausted additionally to maintain the same current when back-emf is induced by a magnetic field?

I think that would be way over my head. You would need to factor in the rate of change, and figure out what the equivalent resistance is of the circuit including the power supply to determine how much of a difference would actually be made.

 

[Arouse1973]

Ok XRZ lets have a go. You need gives us some info. We need to know the size and length of the wire and current in the wire. We need to know what the EMF source is. Inductance, frequency, and voltage. Make it up if you have to but be reasonable. If not I can assume a few things. I cant do it right now as I am on my Tablet.
Adam.

 

[XRZ]

Okay, consider it to be a copper square that is 20 cm wide, 0.4cm thick, 20 cm long. Its resistance is about 4.2x10^-5 Ω, inductance : 56 nH , magnetic field is a 0.5T uniform field from an electromagnet.
Voltage should be: 0.0000042Ω x 1000A = 0.0042VoltsDC. Imagine the magnetic field changing contniously(however you would like to imagine it going on/off or rotating next to conductor). However, the change is massive. delta(BA)/delta(t) = 0.5Tx0.4m^2/0.100s = -2VDC induced back-emf.

What is the factor of power jump here? From 0.0042VDC x 1000A = 4.2W to what...?

I hope this is a reasonable example, if you need to please adjust it! for frequency I dont know why it's needed since it's a DC circuit?

 

[Arouse1973]

Thats quite precise lol is this a question from a book?
Adam

 

[XRZ]

The intuitive power required to do this: (2.0042VDC)^2/(0.0000042Ω) = 956385W, That's 227710 times greater than the original input.

 

[XRZ]

Nope, just made it up @Arouse1973 haha.

 

[Arouse1973]

Yeah right.

 

[XRZ]

@Arouse1973 I really did... I've had this idea in my mind for a while now so...

 

[Arouse1973]

Ok whats the project or is it secret?

 

[XRZ]

Its not a project at all...an experiment to prove the possibility of maintaining the same current on a conductor, without exhausting enormous power.
I self-teach myself electronics and tropics in electrical engineering on my own free time(oh and Physics!), and really tend to think of abstract experiments!
Which teaches me a lot of things, and sometimes conceptualize more.

 

[XRZ]

@Arouse1973 it seems that I caught your interest?
Btw, is what I'm proposing common... or I'm asking something weird...? Because I don't think this is a typical question/problem.

 

[Arouse1973]

Its not a project at all...an experiment to prove the possibility of maintaining the same current on a conductor, without exhausting enormous power.
I self-teach myself electronics and tropics in electrical engineering on my own free time(oh and Physics!), and really tend to think of abstract experiments!

Thats good I am the same. I am not academic so my brain is free to think way outside the box at times. I love Physics also and how odd it can be at times
Adam

 

[Arouse1973]

@Arouse1973 it seems that I caught your interest?
Btw, is what I'm proposing common... or I'm asking something weird...?

Its not common but interesting none the less.
Adam