Resistors and load ratings

[StormCrow]

Hi guys

Bit of a basic question for anyone in the know, but I'm not, so here goes.

I'm replacing four 5w 501 bulbs in one of the cars with 1w LEDs.

Unfortunately, I need to wire a resistor in to the circuit to stop the bulb warning lamp being illuminated.

Bear with me - correct me if I'm wrong:

4*5w = 20w 'normal' draw
4*1w = 4w draw with new bulbs
Need to simulate 16w of draw.

Now here is where my knowledge of electronics gets a bit hazy - can I just fit a 16w resistor (which presumably would be fucking massive?) or am I looking at a 10ohm resistor?

[oldpink]

V= I x R or any variation of the formula

https://people.usd.edu/~schieber/psyc770/resistors/ohms4beginner.html

look at the 3rd example "Let's say a friend asks you to fit a red warning light into the dashboard of his/her car"
same maths
____________________
I have become comfortably numb

Theory & hazard 24-may 2016, CBT 8th June 2016, MOD 1 2nd Aug 2016 Mod 2 2nd-Nov 2016 - Current bike CBR 600 RR

[cimbian]

You will need to know the current drawn by your LEDs and the voltage dropped across them (probably have built-in current limiting resistors so will likely be battery voltage)

Deduct the current drawn from the overall current in your 16W circuit and it is that current you need to shunt through your parallel load resistor.

Use the info on the page OldPink posted and you will find the answer.

Post what you calculate on here if you want someone to check your workings before spend dosh.
____________________
22PlusY
Current: Moto Guzzi 1100 Breva. Previous: Honda XL650V TransAlp
Bearded, Balding, Born again Buddhist Biker

[Rogerborg]

This is what electrical tape was invented for: covering unwanted warning lamp bulbs.
____________________
Biking is 1/20th as dangerous as horse riding.
GONE: HN125-8, LF-250B, GPz 305, GPZ 500S, Burgman 400 // RIDING: F650GS (800 twin), Royal Enfield Bullet Electra 500 AVL, Ninja 250R because racebike

[Islander]

A 10 ohm resistor will draw 1.2A giving you a tad over 14W which should be sufficient to do the job. It'll need to go in parallel with the existing loads. I'd use a 20W type to give you some headroom and remember it'll get hot so you'll need to be careful where you physically place it.

ETA

It might be easier to fit four resistors, one in parallel with each of the LEDs. In which case you'd need 4 x 33 ohm resistors or you could use 39 ohm resistors. Either would need to be rated at 5W minimum and again would need careful placement as they'll get hot.

[oldpink]

this type is good for heat dissipation

https://meniscusaudio.com/images/10R1.jpg
____________________
I have become comfortably numb

Theory & hazard 24-may 2016, CBT 8th June 2016, MOD 1 2nd Aug 2016 Mod 2 2nd-Nov 2016 - Current bike CBR 600 RR

[Islander]

Yep wirewound ceramics are good, easy to find and also nice and cheap.

This type is more expensive but can be fixed to a bulkhead easily - you can even use a smear of heatsink compound under it to help with dissipation.

https://i01.i.aliimg.com/wsphoto/v0/674625655/10pcs-6-Ohm-6R-50W-Watt-Power-Metal-Shell-Case-font-b-Wirewound-b-font-font.jpg

[cimbian]

Given away the answers.. life it just too easy.
____________________
22PlusY
Current: Moto Guzzi 1100 Breva. Previous: Honda XL650V TransAlp
Bearded, Balding, Born again Buddhist Biker

[chris-red]

Why not just get a flasher unit to suit? Might be more expensive but Will be much neater.
____________________
Well, you know what they say. If you want to save the world, you have to push a few old ladies down the stairs.
Skudd:- Perhaps she just thinks you are a window licker and is being nice just in case she becomes another Jill Dando.
WANTED:- Fujinon (Fuji) M42 (Screw on) lenses, let me know if you have anything.

[Islander]

[chris-red]

[smegballs]

Lets say you get two of these in parallel: https://spiratronics.com/10w-alum-clad-wirewound-resistor-22r.html

That will give you a resistance of 11 Ohms.

Assuming a voltage of 12V

I = V/R = 12/11 = 1.09 A

P = V x I = 12 x 1.09 = 13W

Assuming a Voltage of 13.5V

I = V/R = 13.5/11 = 1.22 A

P = V x I = 13.5 x 1.22 = 16.5W

So you should be looking at the right kind of ballpark figure. The problem with a resistive load, as shown is that the actual power drawn varies with the voltage.

In any case, the power rating is being shared by two equal value 10W resistors in parallel. This means they both share the current which means the total allowable power is 20W. So with normal battery voltages you should be well within safe limits.