LED Game Cartridge

[Gaara Ichimaru]

Thanks.

I thought that Resistors only dropped the Voltage, not the Amps.

And I thought that Thin Wire would decrease the resistance...

But I was incorrect there!


And of course, you would put thicker wire where the Main Power is


Thank You.

I'd like to refer you to my essay/metaphor on it.

And that is why a resistor is called a resistor, current is the flow of the electrical particles, think of it as a hall way,
the wire is a hallway, and the current is the people all crammed in the hallway trying to run through it, well the resistor basically narrows the hallway, making it harder for the people (Current) to get through, therefore less people are able to get through and It takes longer for them to reach their destination (The LED). The less people passing the destination (The LED) the less power it gets And voila! my essay on current and resistance is over

[mootoo]

i=u/r => u=ixr => r=u/i when u have 5v on bat and U use led witch need 20mA u need 5v/0.02A = 250Ohm than if LED is 3.6V&20mA than u can find 3.6v on LED and 1.4V on R. R talk about what current can go/flow thru him ,if U change r to 300Ohm than i=5v/300 i=0.01666 and LED is not so bright cause u=3.6V and i=aprox17mA than power of ur light is (in theory) P=UxI P=0.0612W=61.2mW with 250Ohm is it 72mW . If U have two r , u is same i is same but u on each R is something else , so if R1 is 10Ohms and R2 is 20Ohms all circuit have 30Ohms, if u=10v => i=10/30=0.33 so onR1 is 0.33x10=3.3v and on R2 is 0.33x20=6.6v . sorry for my bad "english" but all clear?

[Gaara Ichimaru]

i=u/r => u=ixr => r=u/i when u have 5v on bat and U use led witch need 20mA u need 5v/0.02A = 250Ohm than if LED is 3.6V&20mA than u can find 3.6v on LED and 1.4V on R. R talk about what current can go/flow thru him ,if U change r to 300Ohm than i=5v/300 i=0.01666 and LED is not so bright cause u=3.6V and i=aprox17mA than power of ur light is (in theory) P=UxI P=0.0612W=61.2mW with 250Ohm is it 72mW . If U have two r , u is same i is same but u on each R is something else , so if R1 is 10Ohms and R2 is 20Ohms all circuit have 30Ohms, if u=10v => i=10/30=0.33 so onR1 is 0.33x10=3.3v and on R2 is 0.33x20=6.6v . sorry for my bad "english" but all clear?

I'm sorry, I'm a bit of a newb and I didn't get a word but if you go here it becomes easy 'http://led.linear1.org/1led.wiz'
Thanks though

-Gaara

[mootoo]

i = current in Amp.
u=( idk in english) voltage in Volts
r= resistance in Ohms

[Gaara Ichimaru]

Here's the final product!



And again!



And here is the video evidence

http://www.youtube.com/watch?v=-s302GT40tU

Yay!

First mod -Finished!

Thanks for all your help, in particular Lucid for answering my insane amount of questions and haunter for offering great alternatives.

-Gaara

[lucidPerspective]

looking pretty good! glad I could help out