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tpworks
04-26-2004, 03:52 PM
Say I need 20 ohms and I only have 10 ohm resistors, by connecting 2) 10's end to end what is the effect? Is it going to multiply the resistance by 10 ? and also if I connect them parallel does it just increase the wattage?

ToyMaker
04-26-2004, 04:15 PM
Resistors in series add resistance, so two 10s will make a 20.
Two *equal* resistors in parallel halve the resistance, so two 10s will make a 5 at double the power handling.

robotic regards,

Tom
= = = = =
Times are fun when you're having flies.
Kermit The Frog

Ferenczyg
04-26-2004, 05:41 PM
If two resistors (R1, R2) in series, the total resistance global is the sum of the resistances: R'=R2+R1

If two resistors (R1, R2) in parallel, the total resistance is R'=(R1*R2)/(R1+R2).

As Toymaker said, if the two resistors are equal (value R, for example) and you wire it in parallel then the final resistance is R'=(R*R)/(2R)=R/2

Fer

ESjaavik
04-26-2004, 06:28 PM
Connecting them end to end will as told above give you 20ohm. The wattage will be shared by the two, so it will double.

tpworks
04-26-2004, 06:29 PM
OK so I have only 10 ohm resisters in 50 and 25 watt, in order to get 16.5 ohms at 122 watts min.
is this diagram correct?

50 - 50 - 25 parallel
50 - 50 - 25
50 - 50 - 25
50 - 50 - 25
50 - 50 - 25
series
= 16.5 ohms at 125 watts
or is it 625 watts

50-50 parallel
50-50
50-50
series
does this =15 ohms @ 300 watts

Ferenczyg
04-26-2004, 09:42 PM
Take // as the simbol of parallel, and + as the symbol of series, so:

10//10=5
10//10//10//10=2.5
10//10//10//10//10//10//10//10=1.25
10+(10//10)+(10//10//10//10//10//10//10//10)=10+5+1.25=16.25 ohm

Do not worry about the 0.25 of difference, the resistors usually have a 5% or 10% tolerance.

With the array of 11 (eleven!) :eek: resistors if you use 25W you have 275 Watts.

But with the size of the power resistors this assembly will look like a eiffel tower and can be used to warm your bedroom :cool: .

Honeslty, I think you must consider buying new resistors or some other approach

Fer

tpworks
04-26-2004, 10:41 PM
the reason I am using these figures is I recently purchased a large amount of these resistors on Ebay.

Here is the description.

Large Lot of Misc Dale Resistors
Large lot of Dale resistors inclludes (37) RH-50, 50 watt, 10 ohm, 1% resistors and (94) RH-25, 25 watt, 10 ohm, 1% resistors. Also included are (10) IRC SXAL50, 10 ohm, 5% resistors.

10//10+10//10+10//10=15ohm
2)25w 2)25w 2)25w = 150w
will this work using 6 per motor winding

tpworks
04-26-2004, 11:05 PM
or using 1 set of 5 resistors per motor will this work

10 // 10 =5
10 // 10 // 10 =3.33
5 + 3.33 = 8.33
and all 50 watt does it = 250 watt

balsaman
04-26-2004, 11:17 PM
I would go option 1, 6 resistors per coil. Get a big heatsink. Very big. You will be dumping 250 watts per motor into it. That is a whack of heat. What board will you use?

Eric

tpworks
04-26-2004, 11:22 PM
It's a home brew using 5804 to fire irf540's

Ferenczyg
04-27-2004, 12:34 AM
You really need to put 48V to those 3.27V coils? I know that 8x to 10x the faceplate are frequent, but 15x starts to sound a little heavy ;)

Only moving to 24V for example will give you plenty of speed and you need only

ESjaavik
04-27-2004, 01:09 PM
How does that calculator find 2,7A through an 8,3ohm resistor to give 242,86W?

I thought you would have max 37,25V over the 8,33ohm creating 166,6W total heat carrying 4,472A

Ferenczyg
04-27-2004, 01:46 PM
The developer is three posts above yours ;)

Fer

balsaman
04-27-2004, 05:49 PM
P=ExI

the voltage is 44.73 volts (48 - 3.27) and the since he selected one resistor the current is 2.7 x 2 = 5.4 amps

44.73 v x 5.4 a = 241.54 W

tpworks
04-27-2004, 07:54 PM
I think I finally came to a conclusion to this resistor matter,
10 // 10 + 10 // 10 = 10 ohms @ 200 watt if I use 50 watters
or 100 watts if I use the 25's
Thanks for the help guys.