# LED Lights

Hi

Can anyone out there help . I have a number of 12volt x 4 led low wattage clusters at my disposal. These already have a resistor that drops 12v dc input to the clusters 6v working voltage of the 4 LED's. I would like to use these instead of the 36v lamps ( I don't want to fit a dc/dc converter / tap off one of the 12 volt batteries or re-wire the loom / switches etc ) . Can anyone tell me what value of resistor I need to drop the 36v scooter voltage to 12volt LED ??

Thanks,

Wuffa.

You have to determine the current the LED's draw at their 6 volt working voltage (separate subject). Then use Ohm Law E=IR, where E is voltage I is current and R is resistance.

Example: if the LED's draw, say, 300 milliamperes and you need to drop the voltage from 36 to 6 volts 30 volts: R=E/I or R=30/.3 = 100 ohm.

Let value of present 12 V to 6 V resistor = R, and new resistor = Rx:

R/12 = Rx/36

Rx = [36 R]/12

Rx = [3][R]

----------------------

That Rx will drop the 36 V to 12 V.

To replace existing R, with a new single Resistor (Rs):

Rs = R + Rx

For Power of Rs:

P = (E)(E)/Rs = 1296/Rs

{Note: Common practice is to double the calculated value of P, to allow a safe margin. If P = ~1 W, select a 2 W resistor}

-----------------------------

If the existing R is sealed, and value unknown, just apply 12 V, and measure I.

R = E/I Here, E = 30 V [for the additional drop required]

Select the next higher standard value of R, which will give a slight decrease in light, & a slight decrease resistor load.

Let value of present 12 V to 6 V resistor = R, and new resistor = Rx:

R/12 = Rx/36

Rx = [36 R]/12

Rx = [3][R]

----------------------

That Rx will drop the 36 V to 12 V.

To replace existing R, with a new single Resistor (Rs):

Rs = R + Rx

For Power of Rs:

P = (E)(E)/Rs = 1296/Rs

{Note: Common practice is to double the calculated value of P, to allow a safe margin. If P = ~1 W, select a 2 W resistor}

-----------------------------

If the existing R is sealed, and value unknown, just apply 12 V, and measure I.

R = E/I

Here, E = 30 V [for the additional drop required]

Select the next higher standard value of R, which will give a slight decrease in light, & a slight decrease resistor load.

Let value of present 12 V to 6 V resistor = R, and new resistor = Rx:

R/12 = Rx/36

Rx = [36 R]/12

Rx = [3][R]

----------------------

That Rx will drop the 36 V to 12 V.

To replace existing R, with a new single Resistor (Rs):

Rs = R + Rx

For Power of Rs:

P = (E)(E)/Rs = 1296/Rs

{Note: Common practice is to double the calculated value of P, to allow a safe margin. If P = ~1 W, select a 2 W resistor}

-----------------------------

If the existing R is sealed, and value unknown, just apply 12 V, and measure I.

R = E/I

Here, E = 30 V [for the additional drop required]

Select the next higher standard value of R, which will give a slight decrease in light, & a slight decrease resistor load.

That's right, I should have read the problem more carefully.

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