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Electromagnet V vs Amp vs Turns?

supak111

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Apr 29, 2012
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Hey guys just a quick question, I understand that to make a stronger electromagnet you need to increase the turns, or voltage or increases wire size for more current.

My question,: to turn a 10 watt DC magnet into say a 20 watt DC which of these in the end would make a stronger magnet:

To simple double the voltage? Or change wire gauge and/or turns?

Just wondering which would have less losses/more power?
 

Bluejets

Oct 5, 2014
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For basics, one could say 10W =10V * 1A
Resistance of that circuit would be R=E/I = 10/1 = 10R.
Take that 10R and apply 20volts
I=E/R = 20/10 = 2Amps
Watts = VA in DC so 20 * 2 is now 40 Watts.
 

supak111

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Wouldn't the resistance stay the same at 10R since I didn't change the turns and wire gauge in the coil?
 

supak111

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Oh sorry I miss understood that. R did stay the same "I" didn't. So basically only way to make it "20v" and 20w would be to increase the R as well to 20R with more turns or thinner wire.

Still my question remains... Whats better for an electromagnet, higher V and R or lower V and R?

say 100v 1000R == 10W?
or 10v 10R == 10W?

cost of component not considered, just whats more powerful and efficient
 

Bluejets

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10 watts is 10 watts .
That's the power of each.
Same...same.
 

supak111

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So electromagnets don't care for higher current over higher voltage or vise versa?
 

Harald Kapp

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It's the current that generates the magentic field, not the voltage. The relationship between current and voltage is due to the ohmic resistance of the coil (for DC current, that is).: I = V/R (Ohms law).
The power is given by: P = I * V

Therefore: P = (V/R) * V = V² / R

From this you see that you need to adjust the resistance of the coil by adding or removing windings or using different size wire to match the power to the voltage applied.

The strength of the magnetic field in an electromagnet is in a first approximation proportional to the number N of turns and the current I through the coil:
H ~ N* I
The voltage required to drive the current is a by-product of the resistance of the coil. In an ideal conductor (R = 0 Ω - a supeconductor) no voltage is required to drive the current. By using thicker wire (less resistance) you can create a magnetic field of the same strenght as with a thinner wire but at a lower voltage, hence lower electrical power.
 

Alec_t

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..... but if the electromagnet space is restricted, then using thicker wire means you can fit fewer turns into that space, so you need to increase the current to maintain the field strength.
 

duke37

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The field will be proportional to the ampere-turns and will need a certain power. Fill the spool to minimise the power.
The limit will be due to heat generation, a short term overvoltage can be used if necessary.
 

supak111

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Ok I got what you guys are putting down now. Its not the V that is the question here, its the R depending on turns and gauge of the wire that dictates the whole thing.
 

Bluejets

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10W = 10Volts * 1 Amp
Resistance therefore is E/I = 10V/1A =10R

Same 10R coil

20W = V squared /R
20 = Vsquared / 10R
transposing.........
200 = V squared
so square root 200 = V

V = 14.142 volts

Point being, one does not double the voltage to get double the power.
 
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