Question about ohm's law??

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nerosrevenge

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If I have an led and a 9v battery, how do I calculate the value of the resistor required to operate the led using ohm's law?

I assume I will need some additional values in order to complete the calculation but I am unable to locate these values.

I know that I=V/R.

I think (if I remember correctly) that led's require 6v (or is it 3v??).

So I need to drop 3v across the resistor, this is where I'm stuck.

If I rearange the formula (R= V*I) then I require the value for "I" (current). How do I calculate this value?

Thanks!

Steve.
 
I think the Equasion for caculatic LED resistors is:
(V {Batery} - V{LED}) / I {LED} = R

V {Batery} is the batery voletge
V{LED} is the LEDs forward voltege
I {LED} is the LEDs nomal curent (in amps,NOT miliamps)
R is the resistance reqierd (in Ohms)

For the "I {LED}" use a litle les the the maximum cntinues curent of the LED or the LED might overheat (burning out the LED) on long operatin
 
you should consider the led as a current operated device. Then decide how much current you want. For a normal red led it is somewhere around 20mA. The forward voltage on a normal red led is 2.1V, so:

(9V - 2.1V)/20mA = 345 ohms

In practice, you could use 330, 350, etc, they are all close enough. You can check the dimensionality too to make sure you are doing the right thing.
 
Thanks for the replies. Let's say that I want to maximise the life of the battery. How low can I go on the current?
Then decide how much current you want. For a normal red led it is somewhere around 20mA
I am assuming here but using less current should extend the life of the battery yes?
 

Yes, the lower the current the longer the battery life. It all depends on how bright you want the LED, the more current the brighter it is - so by reducing the current you reduce the brightness accordingly. You need to try higher value resistors until it's just bright enough for what you want.
 
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