View Full Version : Power resistors in series

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.

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