LED indicators - resistor in series or parrallel?

LED indicators - resistor in series or parrallel?

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Discussion

Benbay001

Original Poster:

5,801 posts

158 months

Wednesday 13th April 2016
quotequote all
Hi guys.

I was outside this afternoon working on my bike. The bike has LED indicators fitted by previous owner and the flash rate has always been wrong.

I was fitting an electronic relay.

The other day when probing with my meter i got a volts reading of 12 going into the old relay and 10 coming out, to the indicators, strange i thought but i assumed it was an anomoly.

Then today, when i removed the insulation tape on loom i came accross an 800 ohms resistor crudely soldered in series with the cable running to my indicators.

Is this wrong? It should be wired in parrallel to increase the current, shouldnt it?

Please confirm/ deny my knowledge as its bugging me that im wrong.

Anyway i desoldered it out and stuck my new relay in and all works well so im a happy chappy!

Cheers
Ben

Quinten

1,142 posts

242 months

Wednesday 13th April 2016
quotequote all
You are correct, the resistor should be parallel, to increase the load, not to decrease the current

Benbay001

Original Poster:

5,801 posts

158 months

Wednesday 13th April 2016
quotequote all
Quinten said:
You are correct, the resistor should be parallel, to increase the load, not to decrease the current
tyvm smile

McVities

354 posts

199 months

Friday 15th April 2016
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Isn't resistance in parallel less than resistance in series? confused

Benbay001

Original Poster:

5,801 posts

158 months

Saturday 16th April 2016
quotequote all
McVities said:
Isn't resistance in parallel less than resistance in series? confused
As far as i am aware, resistance is resistance. Ohms doesnt change.
Impedence may change, but i dont have a fking clue what impedence is. biggrin

For all of your informations, i put the bike back together today and the indicators flash beautifully.

Mr2Mike

20,143 posts

256 months

Saturday 16th April 2016
quotequote all
McVities said:
Isn't resistance in parallel less than resistance in series? confused
Yes, if you put resistors in parallel the total resistance reduces, and consequently the current increases.

stocky_m

6 posts

97 months

Sunday 17th April 2016
quotequote all
A conventional flasher unit flashes fast if one of the bulbs is blown and open circuit (high resistance).

The resistance of an LED bulb is much higher than that of a normal bulb so a conventional flasher unit with LED bulbs flashes too fast.
Hence the need to reduce the resistance of the circuit so that the flasher unit flashes at the normal rate.

So, resistor goes in parallel with LED bulbs.

CrutyRammers

13,735 posts

199 months

Sunday 17th April 2016
quotequote all
...but then if you're fitting an electronic flasher relay, you don't need the resistor at all. The electronic relays don't rely on the external resistance to drive the flash rate.

Auntieroll

543 posts

185 months

Monday 18th April 2016
quotequote all
If the resistor is soldered in parallel to the feed wire it is effectively short circuited , hence it is doing nothing as the current will be taking the path of least resistance.

Mr2Mike

20,143 posts

256 months

Monday 18th April 2016
quotequote all
Auntieroll said:
If the resistor is soldered in parallel to the feed wire it is effectively short circuited , hence it is doing nothing as the current will be taking the path of least resistance.
The resistor should be connected in parallel with the LED indicator. Connecting a resistor across two points of a short bit of wire is nonsensical, why would anyone do that?

A proper electronic indicator relay is a better and neater solution though.

Edited by Mr2Mike on Monday 18th April 10:38

Benbay001

Original Poster:

5,801 posts

158 months

Monday 18th April 2016
quotequote all
Just to clarify, i took the resistor out, put the electronic relay in and it works a treat!

Markbarry1977

4,077 posts

104 months

Saturday 23rd April 2016
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Series

R1+R2+R.........=RTotal

Parallel

(R1XR2XR......)/(R1+R2+R......)=RTotal


V=I*R

I=V/R

Hope that helps with your understanding