So if i have a red led that requires 2volts of forward input and i want to give it 10mA i use a 300 ohm resistor since (5v-2v)/10mA = 300.
So my question is, doesn't the led give off resistance? there is of course a voltage difference on it so why isnt that accounted into the calculation. I'm just wondering because if that's the case then it really isn't 10mA going through the led is it? why is the one resistor, in this case 300 ohms, determining the current when the led consumes 2 volts and therefore isnt really that negligible? this would be a lot easier if my amp meter worked but it really doesnt so can someone explain ( and yes im connecting my amp meter in series).
If the LED requires 10A then your calculations are fine and the LED will be getting almost 10A, if you then put in another LED in series (or parallel - reasons explained below) then it would not be getting quite 10A.
The voltage from the battery (i presume a battery) drops slightly when under load (unless it is fully charged and new, then it is probably just over the said voltage) so even with a single LED, it will not be getting quite 10A but I doubt that matters in your situation...
Mowcius
doesn't the led give off resistance?
Give off resistance is an odd phrase. But the concept of resistance is only valid in liner devices and an LED is a non liner device.
there is of course a voltage difference on it so why isnt that accounted into the calculation.
It is taken into account, it is the reduction in voltage. Once an LED has over 2V across it there is nothing to limit the current, it's like turning on a switch. Then the only thing to limit the current is the resistor. As the resistor and LED are in series the same current flows through both.
EDIT
If the LED requires 10A
WOW what an LED!!!!!!
I feel I could do better than my first post.
First off this is not a stupid question, it's a fundamental one not a stupid one.
First of lets see what ohms law actually says:-
"The voltage across something conducting electricity is proportional to the current through it."
so V is proportional to I
Like all math we can make the "proportion" bit into an equals if we put in a constant of proportionality, like this:-
V = k * I
The constant of proportionality is called "resistance" and so we write :-
V = R * I
Now consider a substance where "The voltage across something conducting electricity is NOT proportional to the current through it." This is called a non liner substance. Take for example the filament in a flashlight bulb. When it is cold it has a low resistance but when it is glowing white hot it has a high resistance. The resistance is now some function of the current. We could write:-
V = f(I) * I
In words this is the voltage across it is equal to some function of the current times the current. The resistance has dropped out. Of course you could just write:-
V = R * f(I) * I
In effect extracting a constant out of the function. In the case of a flash light bulb we can do that and quote the resistance at various currents or temperatures.
Now an LED is a device the ideally has no current flowing through it, that is it looks like an infinite resistance, when the voltage across it is below it's "turn on voltage". When the voltage across it reaches this point it all of a sudden has no resistance or looks like a short circuit. So how do we deal with that with ohms law?
Well we can't. So instead we say that if we have a resistor and LED in series we know that the same current flows through both devices. The voltage across the resistor will be proportional to the current through it AND the current through the LED is the same as the current through the resistor. We know the LED will have it's turn on voltage across it so we can express ohms law for the resistor as:-
V_total - Vled_turn_on = R * I
So it would be wrong to think of the LED as having a resistance that is independent of the resistor. The LED does have and "equivalent" resistance but it's value is wholly dependent on the value of the external resistor (providing the voltage across the two is greater than the turn on voltage of the LED).
In practice the voltage / current graph of an LED is not a right angled curve but something a bit smoother but this is normally ignored when calculating current limiting resistors as this is a secondary effect.
I hope that makes things clearer.
dude, absolutely awesome. Thanks for the detailed reply.
Can I just add my thanks to Grumpy_Mike for that explanation - that's the first time someone's gone into the whole current limiting resistor thing and I've understood it. At least I think I do...
Andrew
Wow. That is a really good explanation Mike. I almost understand it now too. One question, you refer to 'non liner' devices. I've never heard that term. You don't mean nonlinear devices do you? Is that an alternate or misspelling or does it mean something entirely different.
Thanks again for the great explanation.
I would think it's a mikes misspelling of linear. But its fun looking on Wikipedia what is the definition of linear is.
I would think it's a mike`s misspelling of linear.
Yes spot on.
I am dyslexic but at school it was called being bloody thick. 
By the way I am dyslexic too, so my idea is can't you be smart at every thing? , we will just have to be satisfied with been brilliant at every thing else but spelling.
Being dyslexic I can tell this joke “Have you have the joke about the dyslexic who could not sleep at nights wondering if they were a one true “dog”
When I was a lecturer I was asked by my head of department to give some advice to a student who was dyslexic. My advice was:-
Never play Scrabble for money
^ lol ;D
I am dyslexic but at school it was called being bloody thick.
Oh, so you're OLD, too... 
(me too. Older than average, anyway.)
As the song does “you are only as old as the woman you fell”
Is that a spelling joke?
I haven't felled any women, trees yes women no.
I'm envious of someone who has dailysex.