Two resistors in parallel. Ohms cut in half?

[Orion]

If you put two resistors together in parallel, are their ohms, for each, cut in half?

To restate, if you have two 1 watt, 4 ohm resistors, put them in parallel, do they now have a 2 ohm resistance each, or would I then have a 2 watt, 2 ohm resistor?

Thanks!

 

[MoonRise]

No.

Two 4 ohm resistors in parallel have an equivalent circuit resistance of 2 ohms.

formula for parallel resistors:

1/R(t) = 1/R1 + 1/R2 + 1/R3 ...

where R(t) = equivalent TOTAL resistance

So, 1/4 + 1/4 = 1/2 so R(t)=2 ohms

 

[Orion]

I see. If both are 1 watt resistors, do two in parallel make 2 watts of resistance?

 

[evan9162]

Depends. If you twist the leads so they are very close together, your power dissipation capabiltiy goes down, so 2 1W resistors very close together might only safely dissipate 1.5W. If they are spaced apart a bit, and have some airflow/airspace, then you should be equivalent to a 2W resistor.

One thing to note, it's best not to exceed 1/2 of a resistor's power rating for safety purposes. So, it's best not to exceed 1/2W of dissipation for a 1W resistor.

 

[Orion]

Maybe I'd better just stick to a 2 watt resistor. Thanks guys!

 

[Lynx_Arc]

what are you using this combination for anyway?

 

[Orion]

Well, I already have some 1 watt resistors but needed 2 watt resistor. I'm just going to the electronics store and buy a package. /ubbthreads/images/graemlins/icon3.gif

 

[Orion]

Well, the 2 watt resistor is too thick for what I was wanting to use it for. /ubbthreads/images/graemlins/frown.gif

Should I go with the two 1 watt resistors, . . . . .or maybe three [unless that's overkill].

 

[WildRice]

total resistance it equivelant to the reciprical of the sum of the recipricals. (gotta love the definition).

If the resistors are the save value it is easy.
2ohm + 2ohm = 1 ohm
3ohm + 3ohm + 3ohm = 1 ohm

and so on..

and I will leave it at that.

Jeff

 

[Orion]

Well, crap. This is more involved that I had hoped. I've got two 3.9 ohm 1 watt resistors and a 2.7 ohm 2 watt resistor. The two 1 watt resistors probably won't work becuase of their limitation, and the 2 watt resistor doesn't fit. The 2 watt resistor at 2.7 ohms is the value I need. /ubbthreads/images/graemlins/banghead.gif

I'm beginning to think that I'm not going to be able to do this. TWOL, two CR123 batteries, . . . . . . . anyone have a VWOT? /ubbthreads/images/graemlins/icon3.gif /ubbthreads/images/graemlins/jpshakehead.gif

 

[Lynx_Arc]

so what value of resistance are you looking for... rangewise?
the 2 3.9ohm in parallel should be a little under 2ohms I would think so you need 2.2-2.5 ohms instead?

 

[Orion]

I agree that two 3.9 ohm resisters in parallel won't be enough. I'll probably have to go find some resistors in the low 5's and parallel them together. I considered two 4.7 ohms, but even a little more would be better. I don't want to push this thing too hard. It's a small sized light that would probably heat up too quickly if I got close to that 1A mark. Probably wanting to have the output of 700 mA. Give or take, depending on what I can find in the electronics section.

 

[WildRice]

just if anyone dosent know th formula out there. I just recientally worked it out.

lets say you have 3 resistors. 10ohm, 22ohm and 47 ohm

10/1 22/1 47/1

take the reciprical of each

1/10 1/22 1/47

get a common denominator

10*22*47 = 10,340

multiply the top of each number by the bottom of the other numbers..
(1/10) 1*22*47 = 1034/10340
(1/22) 1*10*47 = 470/10340
(1/47) 1*10*22 = 220/10340

add them together
1034/10340 + 470/10340 + 220/10340 = 1724/10340

take the reciprical of that.

10340/1724 = 5.99ohm

VERY handy formula, albiet you need a half page of paper to figure anything out, but it works.

Jeff

 

[nikon]

A simpler way to express it would be.....R=AxB/A+B. If resistor A=10ohm and resistor B=22ohm, then 10x22=220 and 10+22=32. So 220/32=6.875ohm.

Add a third resistor 47ohm and 6.875x47=323.125 and 6.875+47=53.875. So 323.125/53.875=5.99ohm.

Remember that the final resistance must always be lower than the smallest resistor.

 

[Orion]

/ubbthreads/images/graemlins/thinking.gif /ubbthreads/images/graemlins/icon3.gif So, . . . . . I think the two 4.7 ohm 1 watt resistors might be best. /ubbthreads/images/graemlins/blush.gif

 

[WildRice]

two 4.7ohm 1W would be about 2.35ohm 1.75W (if mounted together), that should work. I forgot the equasion LONG ago, durring some of the endless hours at work I worked out the previous formula. yours is easier. will it work R=A*B*C/A+B+C ?

Jeff

 

[Orion]

/ubbthreads/images/graemlins/smile.gif I sure do hope so. If all else, I'll mod something with more room and just use the 2 watt resistor. I have a Legend LX that would work well. Thanks for the help, guys!

 

[Darell]

[ QUOTE ]
WildRice said:
it work R=A*B*C/A+B+C ?


[/ QUOTE ]
I don't think so.

Three 30 ohm in parallel gives you 10 ohm resistance (yes?) and the equation above works out to 300. I'm a bit sceptical about my thinking here since it is off by an exact multiple of 10. /ubbthreads/images/graemlins/thinking.gif

It has been WAY to long since I pretended to learn this stuff!

 

[evan9162]

[ QUOTE ]

will it work R=A*B*C/A+B+C


[/ QUOTE ]

No.

The correct expansion for 3 resistors in parallel is:

R1*R2*R3 / (R2*R3 + R1*R3 + R1*R2)

 

[Lynx_Arc]

I would measure the resistance also, it is very rare but once in a great while you get a resistor that is defective and out of the range specified by the % color band on it. I have learned in life not to trust things that I can easily test instead, that way there are less suprises.