Resistor / Watts calculation in parallel

Under the assumption that the rating for the individual resistors remains unchanged when in "parallel", then your written statement is not false.

Something to keep in mind, however, is that the rated dissipation holds for certain standard conditions that may not pertain when you parallel the parts. As an example, significant heat flows out of a lead-mounted resistor to the circuit board, keeping the resistor's temperature lower than if that path did not exist. If you connect two resistors in parallel and mount them in the hole pair intended for a single resistor, the heat flow into the circuit board will not double and there will be a greater temperature rise for the part bodies. And if N paralleled resistors are bunched together, they will not be cooled by air as well as they would be if mounted in the standard fashion.

A stronger statement, assuming you equalize the dissipation in N resistors, (and retaining the "no change in rating" assumption), is that the rated dissipation is multiplied by N.

I would hope so.

Of the 3 posts answering your question I have seen, (by "mike", "Andrew Holme" and "Larry Brasfield"), the answers are all correct given the ambiguity of your question.

I would concur, at this point.

Yes, (under the "unchanged rating" assumption).

That's true, but not does not contradict Andrew's correction.

The second drawing does not state "could be replaced by ...". It seems (to me) to state an equivalence. The rating of the 4 resistor glom is closer to 1W than 1/2 W, under the "unchanged rating" assumption. Under the more realistic "changed mounting changes rating" assumption, the 2nd drawing 1/2W rating could be closer to correct and the 1st drawing deemed wrong.

Im just trying to confirm my understanding of Parallel circuits and Watts calculations when using resistors.

This is not for any project. But sometimes projects call for 1/2 watt resistors. It means I have to run to the store since I don't keep any of them on hand. This would be a way around this.

Are my simple calculations right?

Basically you can double the wattage if you put two or more resistors in parallel. This will of course decrease the R Value.

All Resistors are 100k 1/4 Watt

___ o-|___|-o | |1/2 Watt 50k -----| |--- | ___ | o-|___|-o

___ ___ o-|___|--|___|-o | | | ___ ___ |1/2 Watt 100k o----o-|___|--|___|-o------o

All Resistors are 100k 1/4 Watt (created by AACircuit v1.28.4 beta 13/12/04

--- The total resistance of two resistors in series is the sum of the resistances.

The total resistance of two resistances in parallel is:

R1 R2 Rt = --------- R1 + R2

In your first example you have:

and since you have two resistors, each of which can dissipate 0.25W, the pair can dissipate 0.5W, just like two 100 watt light bulbs dissipate 200 watts when they're both on.

If we label your second example:

___ ___ o-|___|--|___|-o | R1 R2 | | ___ ___ |1/2 Watt 100k o----o-|___|--|___|-o------o R3 R4

Then R1 and R2 in series will be

Rt = R1 + R2 = 200k ohms

and so will R3 and R4 be, so you now have, in effect:

___ o-|___|-o | 200k | | ___ | o----o-|___|-o---o 200k

which, since they're in parallel further reduces to:

___ o----o-|___|-o---o 100k

and since you've got four 1/4 watt resistors in there, they can dissipate 4 * 1.4 watt = 1 watt.

-- John Fields

Well I have two posters with two different answers.

Perhaps I have labeled my drawings poorly and the confusion starts there.

Each of my expamples consists of 100k 1/4 watt Resistors.

1. The first schematic in this configuration could be replaced with a 1/2 watt 50k Resistor.

1. The second drawing could be replaced with a 1/2 Watt 100k Resistor.

Are you saying that I'm wrong on both drawings, or the second drawing and why?

Thanks.

Watts

Thank you Larry.

Yes I understand that you can't keep adding resistors and not take into consideration the ramifications, eg heat and draw on the rest of the circuit. Thanks for the tip.

If anyone cares I found this usage of resistors to increase wattage on a document

Search for the word Even Hotter where it says.

EVEN HOTTER

If an even higher power rating is required then several resistors may be connected in parallel. Ten 5 Watt resistors connected in parallel would give a total power rating of 50 Watts. The resulting resistance will, of course, be a tenth of the individual resistors.

It was from there that I thought mmm I wonder if I can do the same when I get stuck.

Regards

Novice

Correct.

No. This is 1 Watt 100k

there.

and

I see. I neglected to take into consideration the series resistors. I suppose then that two 1/4 watt 100k resistors equals a 200k 1/2 watt.

thanks.