reduce 21w to 5w?

Hello, i want to use a bulb with 68 smd's to replace a standard filament daytime running light bulb, i am currently bridging the positive and negative with a 470 ohm 1/4 watt resistor, it is working but after only a few days use it looks like flux is being pushed out of all the connection, the bulb smells off burning flux and the resistor is discolouring. To me this means i need a lower ohm resistor to let more power waste before it gets to the bulb. Any idea what size resistor i need or have any other input? 12v 21w feed. I dont have any technical info for the bulb but il try and tell you any other info you need to know. Thanks in advance

470 Ohm across 14.4V is nearly 1/2 Watt (V x V / R) 68 LEDs in a single bulb? Why so many? And how do you intend wiring them - series strings of 3, 4, 5, or 6 etc? With a resistor per string, or one resistor for all, or no resistor?

Thank you for the replies, the bulb is a car DRL (daytime running light) bulb so it needs to be very bright to be seen in the day by other road users. It is a bulb i have brought online with the smds already on it. Its this one - http://www.dreammacys.com/index.php?gOo=goods_details.dwt&goodsid=1022&productname= What i usually do with sidelights is use a 1/4W 470ohm resistor between the + & - legs of the bulb to the full 5W doesnt get to the bulb but obviously now with a 21W feed a 1/4W 470ohm resistor will fry after a short amount of use. So from what IAAI has said (who's name fills me with confidence) i should use 4 1/4W or 2 1/2W resistors to help the power bridge from the + to - better so less power gets to the bulb? I hope this makes the situation a bit clearer, any more help is greatly appreciated!

You don't need resistors - they are 12V lights/LEDs. Different if you were assembling your own LEDs... Remember, loads take what they can. Power cannot be pushed into a load except by increasing the voltage, so such 12V-rated LEDs/lights are fine on any 12V DC supply... (A car's typical ~14V is still "12V" in this respect.)

If you have access to a 470 ohm 1 watt resistor, that will work. If you have (4) 470 ohm 1/4 watt devices, you can wire them in series/parallel to achieve the same 470 ohms, yet the power will be divided equally amongst the 4 parts. If you use (2) half watt devices, you will need (2) 1K devices in parallel, this will yield 500 ohms. Plenty close enough in the LED world. I just noticed that you are paralleling the resistor across the bulb. There is no need for this. As already stated, if the lamps are 12V devices, no need for the resistor. Your house is connected to a line that is capable of providing millions of watts. Your house only takes what is needs. Same in your car.

No! It's a 12V bulb (with lots of SMDs). Note what adam is saying... "What i usually do with sidelights is use a 1/4W 470ohm resistor between the + & - legs of the bulb

... so the full 5W doesnt get to the bulb...

... but obviously now with a 21W feed a 1/4W 470ohm resistor will fry after a short amount of use.". adam - putting a resistor in parallel with the load has NOTHING to do with what the load takes. Remove the resistor on you old setup. or add and remove to the new. Do you notice any brightness change? No - because it has no effect. Do not confuse SERIES resistors for individual LEDs etc - that is to drop 12V to the LED voltages (eg, 470 ohm x 20mA = 9.4V, so the resistor drops 9.4V from normal ~14V for 3.4V LED etc).    You do not need that because those 12V LED "bulbs" are already designed for 12V systems. If you don't believe me, put the LED bulbs or a small 12V bulb (2W, 3W, 5W, or 21W etc) across the car battery. It won't blow despite that battery being able to deliver (say) 12V @ 100A = 1200W. It's all about "matching" voltages...