Quick Check - Re: Paralleling Resistors

[rhd]

Quick check on my knowledge here -

I need a 4W resistor of somewhere around 0.4 to 0.45 Ohms. I don't have such a resistor, but I do have:

2W 1 Ohm
3W 0.75 Ohm

In parallel, they should create 0.429 Ohms of resistance, and roughly speaking, should create a resistor with the ability to distribute that 4W..... However, I know that they won't share the "wattage distribution" (I'm just making up that term) equally, since they are different resistances. IE, one resistor will be doing more work than the other. My concern is around which of the two will be handling *more* of that required 4W total.

My gut / basic logic tells me that the 3W resistor (because it has the lower resistance) is going to be doing more of the work. This would be fine (as long as it's not doing more than 3W) But if it's the other way around, I'll be in trouble.

Thoughts?

 

[Jimmymcjimthejim]

When you put resistors in parallel, the voltage drop across them is the same. In order to compensate for the lower resistance, a higher current will go through. Thus, the power dissipated will be greater in the lower resistance resistor. Hope this helps!

 

[rhd]

I should be fine then, thanks!

I need it to be able to handle 4W... and that should be ok. I don't think the 3W would end up having to dissipate more than 3W in that config... do you?

 

[Cyparagon]

P=V²/R

If 2W is dissipated in the 1Ω, the voltage must be 2=V²/1, V=1.41V.
1.41V across 0.75Ω would be 1.41²/0.75=2.7W, so the 1Ω is the bottleneck
and the equivalent resistor you have is a 4.7W 0.43Ω.

 

[Coherent Light]

rhd try this calculator.

PARALLEL RESISTOR CALCULATOR

 

[HIMNL9]

It depend from the voltage at the leads of the resistor, too ..... supposing you are using it for a LM317 based regulator (1.25V ref), the powers are the following ones:

for the 1 ohm one - 1.25V/1ohm=1.25A -> 1.25V*1.25A=1.57W

for the 0.75 ohm one - 1.25/0.75=1.666A -> 1.25*1.666=2.08W

(approximated to 2 decimals)

If your voltage is different, ofcourse, the values changes.

 

[rhd]

Well - If I'm paralleling a 1Ohm and 0.75Ohm, I'll have a 0.43 Ohm resistor, which should produce a current of ~2.9A
(FYI - It's for an LM338, and for an LED, not an LD - but same idea, same circuit)

So my required wattage would be: 1.25 * 2.9 = 3.6W
But how would that wattage be distributed between the two parallel resistors?

I think the correct math would be:

Wattage @ 1 Ohm Resistor: 1.25^2 / 1 = 1.563 W
Wattage @ 0.75 Ohm Resistor: 1.25^2 / 0.75 = 2.083 W

And those figures sum back to the total 3.6W that would be expected were this to be one resistor. Does everything here look right?