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Quick electrical question


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I need a dummy load unit to get my indicators flashing at the right speed after fitting led rear light units.

But i carn't find a formular anywhere (that's simple enough for me to understand!) to work out what ohms i need.

Something like this is what i'me looking at from RS

Cheers in advance

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Assuming it has to replace 2 bulbs you need a minimum power dissipation of 42 watts, each indicator bulb is 21w.

Use the formual W = (v^2)/R W = power in watts, v = volts and R is the resistance in ohms.

So R = (V^2)/W = 12*12/42 = 3.43 ohms (12 volts). You'll have to choose the nearest preferrfed value, and bear in mind they will get hot so the resistance may alter slightly from the nominal.

Instead of buying a resistor, you could use 2 x 21 watt bulbs, these could be sited conveniently at the front and rear of the vehicle and protected by amber lenses :ph34r:

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Sort of defeats the object of fitting LEDs doesn't it?

Not really, as led's should last much longer, switch on quicker, be more shock resistant and run cooler than filament bulbs :).

If you do the unit wiring right they should be sealed against mud/water, they don't however like dirty power supplies i.e unregulated noisy supplies. :(

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I've used LED lamps myself but TBH I've never found one as bright as a filament bulb. I use them mainly to reduce the current draw in sidelights/brakeleights because I tow quite a lot. If I had to fit a resister to increase current draw then I really don't think I could justify the huge increase in price compared to a filament bulb.##I would agree that a brake light using LEDs can increae safety, provided it's bright enough

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