Electrical and magnetic field question

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Discussion Overview

The discussion revolves around the challenges faced in powering a solenoid coil with an iron core using a variac power supply. Participants explore issues related to electrical resistance, current draw, and the implications of using a variac in this context, focusing on the technical aspects of voltage, current, and power dissipation.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that the resistance of the #8 gauge wire is low, leading to a high current draw at 50 volts, potentially exceeding 100 Amps.
  • Another participant highlights the shock hazard associated with variacs, suggesting the use of a GFCI protected outlet and an isolation transformer.
  • A participant calculates that the power dissipation at 50 volts could exceed the variac's rating, questioning if this is the reason for the fuse blowing.
  • There is a suggestion to add resistors to limit current, which another participant argues against, stating it would be inefficient and that voltage should be controlled instead.
  • Concerns are raised about the solenoid's ability to actuate at high current levels, with a suggestion to increase the number of turns to achieve greater force.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to resolve the issues with the solenoid and variac setup. There is no consensus on whether adding resistors is beneficial, and the discussion remains unresolved regarding the optimal design for the solenoid's functionality.

Contextual Notes

Participants reference specific electrical ratings and calculations, but there are unresolved assumptions about the solenoid's design and the exact requirements for its operation. The discussion includes varying interpretations of the variac's capabilities and safety considerations.

jearls74
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i have posted a few other questions on this topic here and have gotten good answers and advice. i have been working on a project for over a year, i have built a long solenoid coil with an iron core and was trying to power it from a 12 volt car battery with no luck changed my strategy after some advice from a mentor here. he suggested i use a controlled benchtop type of supply, so i have bought a variac power supply, its input is 120 volts ac 50-60 Hz. its output is 130 volts 25 amps maximum. my solenoid is made from #8 gauge enamaled magnet wire. i have used 700 feet. i have connected the output of the variable transformer to a full wave bridge rectifier to convert the ac to dc. the fullwave rectifier is rated for more than 200 volts and 35 amps. my question is this, why does the fuse keep blowing before i get the voltage above 50 volts on the variable transformers voltage gauge?
 
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Well since the resistance of that wire is only 0.00206 Ohms/meter (see http://www.daycounter.com/Calculators/AWG.phtml ), then the total resistance of your coil is only about 0.43 Ohms. At 50 volts DC you could be trying to pull over 100 Amps.

The fact that your solenoid won't engage at that current level may indicate a problem with your mechanical design.
 
Also keep in mind that the output of most variacs is NOT isolated from the AC Mains, so they present a pretty significant shock hazard for ground faults. You should at least connect the variac to a GFCI protected outlet, and better yet, put an isolation transformer between the AC Mains and your variac.
 
good info, uart pointed out the main thing i was doing wrong, at 50 volts with just the .439 ohms of resistance in the coil, the variac was trying to pull over 100 amps and after doing the basic calculation of ohms law, i find that the power dissapation of 50volts x50volts=2500/.439=5694.76 watts, the variac is only rated at 3000VA. does this mean that the variac can dissapate 3000 watts of power itself? is this why i keep blowing the fuse at 50 volts? if so, shouldn't i just add (2) 3 ohm 10 watt resistors after the coil to limit the current and power if the variac unit is dissapating 3000 watts itself? at 130 volts, 6 ohms of resistance, i get 21.66 amps and 2816.66 watts, if the variac does dissapate 3000 watts of power then the wattage of the (2) 3 ohm resistors shouldn't matter? also berkeman enlighten me to something i didnt know about the variac, i didnt know that is was not like a regular transformer where the the primary and secondary coils are not electrically coupled. the fact that the variac is electrically coupled the the ac mains explains why the lights in my place dim right before the fuse in the unit blows. I am going to get an isolation transformer to put in line before the variac. if anybody could answer my question i would appreciate the help.
 
"does this mean that the variac can dissapate 3000 watts of power itself?"

It means that the variac can deliver 3000 Watts without overheating.

"i this why i keep blowing the fuse at 50 volts?"

It means that your fuse isn't rated for 100 Amperes, or whatever uart said you were drawing. That's a lot of amps.

And don't stand in any pools of water while your playing---like I found myself last month...
 
if so, shouldn't i just add (2) 3 ohm 10 watt resistors after the coil to limit the current

No that would be silly and just a waste of power for no gain. You limit the current by limiting the voltage on the variac, that's what it's there for.

You need to find out why 30Amps is not providing enough force to actuate your soleniod, you need to provide some more info about the design if you're are going to figure this part out.

If you really do need more force then you need to add more turns. If your variac reaches maximum rated current at say 20V, then if you triple the number of turns then you can apply 60 volts for the same current and get three times the force. It sounds like you've got a lot of ampere-turns already though, what is this soleniod trying to do anyway?
 
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