
#19
Jan2407, 10:53 AM

PF Gold
P: 1,686

So since you are using the wall outlet, you already know your voltage. I'm assuming that you have a numerical quantity of the magnetic field which you wish to achieve? It depends on how you connect the resistors, in series, the resistances would add. In parallel they would decrease (hover's example). Since in both of these cases, either the voltage drop or current of the resistor changes, the power dissipation could be "spread out" across the resistors. If you use 2 of your 1 mega ohm resistors is series, you would have a total resistance of 2 mega ohms. Knowing that the wall voltage is 120V, we get the total current in the circuit: I = 120/2000000 = 60 micro amps Each resistor would have 60V across it: P = IV P = 60 micro amps * 60 = 3.6 milli watts Your resistors will handle the power dissipation, but your current is small! 



#20
Jan2407, 12:00 PM

P: 338





#21
Jan2507, 10:13 AM

P: 16

I am using a power cord from a regular computer. But that does not allow the whole 120V thru it correct?
If I were to buy different resistors then I would want some that allow less ohms to run thru them? I don't mean to ask stupid questions but I am really confused about some of this stuff because I have never really worked with this technical of stuff on this subject. 



#22
Jan2507, 10:40 AM

PF Gold
P: 1,686

Also I must ask, when you first built your circuit, what was your current through the circuit that caused the wires to melt? 



#23
Jan2507, 01:55 PM

P: 338

i have 2 power cords, one 19volts and the other 16volts. As for resistors get something like this. Resistors like that from radioshack are good for power supplies and come two in a pack. They are also less than $2, you could easily buy it.




#24
Jan2507, 02:05 PM

P: 338

DHS Science i have a question for you, how strong are you trying to make your electromagnet? Are you just trying to run some current through a wire to create a simple electromagnet or are you trying to create the strongest electromagnet you can make??




#25
Jan2507, 08:21 PM

P: 2,449

All I can say is WOW. This should be moved to electrical engineering. There are some things in here that seriously don't make any sense. DHS, you haven't told us what gauge wire you are using in your electromagnet. You are using 1 megohm resistors with a 12 volt battery? There is NO WAY a one megohm series resistor can allow enough current through from a 12 volt battery to melt your wire. You'd have to have 10000 of them in parallel to get down to 100 ohms. There is no way that could even allow enough current through from a 12 volt battery to melt your wire. As for the 15 amps getting through what you guys are calling the 'power cord', well, that IS a power supply. It converts the 120 VAC down to a lower DC voltage. 19 and 16 volts was mentioned. If the thing has a rating of 3 amps, then that is all the current that you will get out of it. If the resistance of the coil calls for more than 3 amps at the power supply's rated voltage, then the power supply will not be able to handle it and its output voltage will drop.
 Hover, you refer to 3 50 ohms resistors in parellel = 16.66 ohms with a total power dissipated in the resistors of 21.66 watts. This is NOT the power in each resistor. In this case each resistor will take one third of the total power of 21.66 watts. You are also assuming zero resistance of the electromagnet coil. We don't know what this resistance is since DHS hasn't told us nor has he told us what kind of wire is being used.  DHS, I'm not sure what kind of wire you are using, but try installing some car headlights in series with your coil. Older sealed beam bulbs are cheap and the ones that have 2 filaments in them can be wired so that both filaments are in series, both in parallel, or just one at a time to get close to the desired current. Obviously a car headlight is able to take the full 12 volts so you can't hurt them. They are simply there to limit the current in your coil so you don't destroy it. 



#26
Jan2507, 08:35 PM

P: 338

Averagesupernova i quote you "Hover, you refer to 3 50 ohms resistors in parellel = 16.66 ohms with a total power dissipated in the resistors of 21.66 watts. This is NOT the power in each resistor.". Are you saying that in order to get the amount of watts in each resistor i do 21.66/3?? 



#27
Jan2507, 08:48 PM

P: 2,449

I am saying that the power in a resistor is equal to the voltage across it squared and then divided by its resistance.




#28
Jan2507, 09:01 PM

P: 338

That way or my way works out exactly although your way is a shorter version. You seemed to miss the point of what i just asked. I said " Are you saying that in order to get the amount of watts in each resistor i do 21.66/3??". I reread your previous post and got this "In this case each resistor will take one third of the total power of 21.66 watts." so i guess that anwsers my question.




#29
Jan2507, 09:03 PM

P: 2,449

I said "in this case" because it only pertains to this case. The safest most sure way of getting the correct answer is E^2/R.




#30
Jan2507, 09:10 PM

P: 338

Just to be clear, say you had 5 resistors i parrallel. Say the amount of power(watts) in the resistors is "X". Would i divide "X" by 5 to get the amount of power in each resistor??




#31
Jan2507, 09:11 PM

P: 2,449

You would only if the 5 resistors are of the same ohm value.




#32
Jan2507, 09:27 PM

P: 338

ok i see now.
Thx 



#33
Jan2507, 09:39 PM

P: 2,449

Let me guess. You did some more math?




#34
Jan2507, 09:44 PM

P: 338

How did you know ? I never put three 50 ohm resistors in series on my electromagnet before because i thought they would all take 21.66 watts of power and overheat.




#35
Jan2607, 10:00 AM

P: 16

Hover:
"DHS Science i have a question for you, how strong are you trying to make your electromagnet? Are you just trying to run some current through a wire to create a simple electromagnet or are you trying to create the strongest electromagnet you can make??" I am trying to make the stongest that i can with the supplies that i have. Averagesupernova: DHS, you haven't told us what gauge wire you are using in your electromagnet. You are using 1 megohm resistors with a 12 volt battery? There is NO WAY a one megohm series resistor can allow enough current through from a 12 volt battery to melt your wire. Sorry about not saying the size of wire, it is 6 gauge. And the reason that the batter melt thru the wire last year was because i never used a resistor then. I just ran a wire straight off of the battery thru the coil. That is the main reason why i have so many questions. Sorry. Last night i went out and bought a pack of 2 resistors that allow 100 ohms and 10W. So if hook both of them up parallel in the line then i should be fine? Because if i do then the power will be divided evenly between the two resistors and will still only allow 10W thru in all, correct? 



#36
Jan2607, 12:24 PM

Sci Advisor
HW Helper
P: 1,572

Power in resistor is E squared times R. In this case 12^2 /100 = 1.4 watts. The resistor does not limit watts. It only provices resistance. #6 wire should be able to handle about 30A to 40A without heating up too much. Current in the coil is E/R. Or in this case 12/50 = 0.24A You may want to use smaller resistance value. For 10W resistors 30 ohms is about as low as you want to go with a 12v source. This provides a safety factor since the 10W rating is only for ideal conditions. 


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