Confused about Electricity

If I have a 9v 200mAh battery. and I put a 10-Ohm resistor on it:

1.) does the current get divided by the 10:
200mAh / 10-Ohm = 20mAh :thinking:

2.) Does anything happen to the voltage? or does it stay at 9v?

3.) With does the 10k-Ohm resistor mean? I don't understand what the 'k' is:thinking:

4.) If I want to get MAX brightness out of an LED, even if the life is decreased a bit, What resistor should I use?
LED Specs:
3v max
20mAh max

Thanks for any help, Electricity is confusing (to me)

Also if would help the matter I have twin to this LED. I'm not against using both if it would help me make it brighter

I have seen things like this before... That still doesn't help. any chance you can answer my questions? I 'did' say, "I'm confused". Meaning I know the material, I just don't understand...

Thanks for trying to help though...

michaeloohra said:

I have seen things like this before... That still doesn't help. any chance you can answer my questions? I 'did' say, "I'm confused". Meaning I know the material, I just don't understand...

Thanks for trying to help though...

Click to expand...

nowhere did you say you knew the material and if you knew the material you would have answered your question. his post contains much knowledge.

V=IR

V= Voltage
I= Current
R= Resistance

simple math can manipulate that formula.

The K means thousand, quite simple.

10k ohm = 10,000 Ohms

Nvm nvm

It you don't figure out how to apply Ohms Law you should stay away from any DIY electronics/electrical projects.

To answer your question:
you deduct the 3V voltage drop over the diode from the total supplied by the battery.
9V - 3V = 6V

Now you want a resistor that can give you a 6V voltage drop at a current of 20mA.
Flip around Ohms Law until you have resistance as a function of voltage and current.
U=R*I > R=U/I
R = 6V/0.020mA = 300 Ohm

Resistors come in certain intervals, the closest to 300 Ohm are 270 and 330.

Go to radioshack or something and buy a 270 Ohm resistor for each of your diodes.
----R--D---
B+< > B-
-----R--D---
Connect them like this.

lol I figured out what I was doing wrong. On one side of this recharagable 9V it says 200Ah, on the other it says 200mAh. I have 2 of the same battery, but only one says it.
So I kept getting anomalies in my math.


BTW one thing you said stuck out. "deduce the 3v drop" I'm assuming that the diode is always going to do this? Thanks.

Yes, diodes have a voltage drop that does not increase much with current.
You better be sure it is 3V, like by measuring the voltage across it when powered up for the first time.

The mAh rating of the batteries is irrelevant for the calculation, it is simply a statement of it's capacity.
A cars meter may go to 160miles/hour, but that does not mean it will go that fast as soon as you turn on the engine.

1) LED's do not have a mAh rating. Ampere Hours (or miliAmpereHours are capacity ratings for storage cells, or Batteries. It means that a particular unit can deliver n mA over a one hour period. That is NOT to say that if you were to short the + & - that the current would top out at n mA. It would, more likely, deliver several Amperes instantaneously and until the delivery mechanics of the battery (chemical reaction, charge depletion, whatever) were overcome by the rapid discharge, or the short melted, ect. A 200AH 9V battery would be one hellashious honker. That must be a missprint on somebody's part. 200mAH seems a little low.
2) Toke gave you a very good, very basic, lesson in series circuit Ohm's Law. You have a series DC circuit (single current path, Direct Current) supplied by 9 Volt source. The total voltage drop accross that circuit will be 9 Volts, some dropped by each circuit element ( resistor and LED) . The Absolute max ratings for your diode are Vf=3Volts at 20mA (MAXIMUM). That says that under MAX conditiond, the LED will drop 3Volts, so the resistor has to drop the other 6Volts. Ohm's law tells us that in a series circuit, each element will pass the same current (20mA at the most). So the resistor has to drop 6Volts (V=6) and pass .02A (I= 20mA= 0.020A). Ohm's Law says that Resistance (R in Ohms) is equal to Voltage (V in Volts) devided by Amperage (I in Amps) hense, R=E/I= 6/0.02= 300 Ohms to avoid exceeding the max ratings of the diode. I would dissagree with Toke, in that I would use a 330 Ohm instead of the 270 to stay within the limits, but with resistor tollerances and manufacturer's tendancy to overstate Absolute Max's, you'll probablly be just fine using a 270 Ohm resistor. It don't get any simpler than that, bud. If you're still confused, take Toke's advice and stay away from DIY electronics. But, Good Luck with your project anyway. Don't parallel the LED's! You'll end up frying one of them, unless you really meant what you said about not caring if you decreased the diode's life (if that is the case, you can get maximum brightness, albeit for a very short time span, by just tying it into the battery without any current limiters). Well, that's about as much yabbeling as I can do. Back to you Toke.

This is what I have done. I still needed a few things to do the suggestions you guys have given me. I get paid tonight so I'll run to radioshack and start the suggested project.

Till then this is what I have, What do you think? is there anyway to make it brighter, or just all around better?

I just uploaded this, so It may take a second to work:
Video of the LED box

I'm interested in suggestions

You missed the point of both my attempt at a drawing in post #7 and 123splat's elaboration.

Your drawing have the LED's in parallel, that means that a slight difference can mean that one gets 10mA and the other 30mA.

You will need two resistors, 330 Ohm is bests.

Place the resistors next to each other and twists one of the ends together, you should get this.
---<===

Connect the twisted end to the battery, connect the two other ends to a LED each, connect the two other ends of the LEDs to the switch.


ETA:
Never mind, after seeing the video with presumable a 10 Ohm resistor, and you wanting them brighter this seems pretty hopeless.

Just to clarify, when you say "twist together" you meaning the bottom diagram, of the pic below right?

Toke said:

ETA:
Never mind, after seeing the video with presumable a 10 Ohm resistor, and you wanting them brighter this seems pretty hopeless.

Click to expand...

Well thanks for your help. I'll keep reading up on this stuff. I didn't know how hard this stuff would be starting out. Perhaps I'm not on the level I thought

Thank you again, for your time and help!

Your resistors should end up Y shaped so that the current splits up, one half for each diode.