# Resistor / Watts calculation in parallel

Discussion in 'Electronic Basics' started by Andrew Holme, Mar 3, 2005.

1. ### Andrew HolmeGuest

No. This is 1 Watt 100k

2. ### mikeGuest

Yes,
If you know the voltage or current, you can calculate the wattage
in each resistor for combinations of different values.

--
Return address is VALID but some sites block emails
..
Wanted, PCMCIA SCSI Card for HP m820 CDRW.
FS 500MHz Tek DSOscilloscope TDS540 Make Offer
Wanted, 12.1" LCD for Gateway Solo 5300. Samsung LT121SU-121
Bunch of stuff For Sale and Wanted at the link below.
MAKE THE OBVIOUS CHANGES TO THE LINK
ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/

3. ### Larry BrasfieldGuest

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.

[Cut some resistor diagrams.]

4. ### Larry BrasfieldGuest

(by "mike", "Andrew Holme" and "Larry Brasfield"),
the answers are all correct given the ambiguity of
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.

5. ### ChretienGuest

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 www.tech-chat.de)

6. ### John FieldsGuest

---
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:

1E5R * 1E5R 10E9R
Rt = ---------------- = ------- = 5E4R = 50K ohms
1E5R + 1E5R 2E5R

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.

7. ### ChretienGuest

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.

2. 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.

8. ### ChretienGuest

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

http://www.sarts.org.sg/lesson2.pdf

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

9. ### Rich GriseGuest

Because 4 * 1/4 = 1.

Cheers!
Rich

10. ### newsGuest

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.  