I'm just trying to play with some leds. It seems the most common resistor for this application was the 220 Ohm resistor so I went to buy some and there is a bunch of different wattage sizes. Nobody ever mentions this.

I have some 150 Ohm resistors. Could these work? I also have some that are 320 Ohms.

I would like to buy a decent parts list for capacitors and mosfets bu t I don't Know what to buy.

So I have $18. How would you spend it? I am going to be hooking up sensors and a recharge shield.

LarryD thank you very much I'll go with the 330s. I cant seem to do the simplest thing without a million details.

I have a lot to learn. Here is your total beginner question of the day. Is the current what the led is rated for, or what flows out of the battery? I don't know where to get the numbers for the equation W= V * C

For normal 20mA LEDs get the lowest wattage and hence the cheapest you can find. That is normally quarter or eighth watt.

The numbers to use are the current through the resistor, which in this case is the same as the LED and the voltage across the resistor which is the supply voltage minus the voltage across the LED.

I have a lot to learn. Here is your total beginner question of the day. Is the current what the led is rated for, or what flows out of the battery? I don’t know where to get the numbers for the equation W= V * C

The current is calculated using [u]Ohm’s Law[/u] (Current = Voltage/Resistance).

The basic power formula and Ohm’s Law can be combined algebraically to get Power = V^{2}/R.

Many LEDs have a max continuous current rating of 20mA.
If the source voltage is 5V, and Vf of the LED is 2.2V, then the resistor to set that up is:
(Vs - Vf)/current = resistor
(5V - 2.2V)/.02A = 140 ohm.
The power dissipated in the resistor is P=IV. V = IR, so sub in: P = I*IR, so .02A * .02A * 140ohm = .056W, 56mW
Alternately, Vr = Vs - Vf = 5V - 2.2V = 2.8V. .02A * 2.8V = 56mW

Say you only had 220 ohm resistors available, how much current would flow?
(Vs - Vf)/resistor = current
(5V - 2.2V)/220 = .0127A, 12.7mA
And .0127 * .0127 * 220 ohm = 35.4mW

CrossRoads:
Many LEDs have a max continuous current rating of 20mA.
If the source voltage is 5V, and Vf of the LED is 2.2V, then the resistor to set that up is:
(Vs - Vf)/current = resistor
(5V - 2.2V)/.02A = 140 ohm.
The power dissipated in the resistor is P=IV. V = IR, so sub in: P = I*IR, so .02A * .02A * 140ohm = .056W, 56mW
Alternately, Vr = Vs - Vf = 5V - 2.2V = 2.8V. .02A * 2.8V = 56mW

Say you only had 220 ohm resistors available, how much current would flow?
(Vs - Vf)/resistor = current
(5V - 2.2V)/220 = .0127A, 12.7mA
And .0127 * .0127 * 220 ohm = 35.4mW

Grumpy_Mike:
For normal 20mA LEDs get the lowest wattage and hence the cheapest you can find. That is normally quarter or eighth watt.

The numbers to use are the current through the resistor, which in this case is the same as the LED and the voltage across the resistor which is the supply voltage minus the voltage across the LED.

Thank you that let's me buy something specific now. I have piles of resistors I'll never use it would seem.

I need to sit down and read a while. That kind of reading is hard for me but I know I got to do it. I have a lot of questions coming Thanks to everyone who contributed an answer.

Paul__B:
Well now, the formula for power is V^{2}/R.

Considering resistors placed directly across 5 V, V^{2} is 25, so a 100 Ohm resistor would dissipate ¼ W, and a 220 Ohm resistor an 11th of a Watt.

Clearly, a tenth Watt rated 220 Ohm resistor is safe in any circuit powered at 5 V.

That’s a great walk through. I have no idea why I was making it so complicated. I mean I hate math, but c’mon. It’s just simple division? Why do they go and make it all complicated looking? I hope there isn’t a ton of math in this. I know it’s hard to be smart without knowing math but I can write code, and I thought that was what I would be doing. Why do I always get sidetracked like this? Just to be safe, I’ll dig out a calculator.

LarryD what do the red dots represent? they look like switches. I usually look up stuff like that but googling “red dots” might not turn out so well.

So I got adventurous and won an auction for $20. How did I do?:
20x Resistor 220Ω
10x Resistor 1kΩ
10x Resistor 10kΩ
3x Potentiometer
1x Photoresistor
4x Red LED
4x Green LED
4x Blue LED
1x RGB LED
1x LED bar graph
1x LED Matrix
4x Push Button
2x Diode
2x NPN Transistor
2x PNP Transistor
1x Active Buzzer
1x Passive Buzzer
1x Relay
1x Motor
1x Servo
1x Joystick
1x LCD1602
1x L293D
1x 74HC595
1x MPU6050 Acceleration sensor
1x LM35 Temperature Sensor
1x Breadboard
65x Jumper Wire M/M
10x Jumper Wire F/M
1x Battery Holder

Understand that if the actual power is 1/10 watt then 1/4 or even 1/2 watt (same resistance) is just fine. I mention it this way because depending on who you are getting parts from, you may find the 1/2 or 1/4 watt resistors (or whatever) are significantly cheaper than say the 1/10 watt resistors. Be sure and check the quantity pricing - often times, 10 or even 100 of the same value are MUCH cheaper than the individual prices. Get common values like 100, 220, 330, 1k, 4.7k and 10k in bulk to have them on hand (the first 3 are typical values for current limiting for LED's etc, the last 3 are more common for pull-up or input protection.

The red dots are just connection points which are joined to other parts in your circuit.
They may or may not have numbers assigned to them for lead reference.

You can think of them as the point where you solder a wire to.

I put the black dots there to say:
"here is where you measure the voltage".

That's a great walk through. I have no idea why I was making it so complicated. I mean I hate math, but c'mon. It's just simple division? Why do they go and make it all complicated looking? I hope there isn't a ton of math in this. I know it's hard to be smart without knowing math but I can write code, and I thought that was what I would be doing. Why do I always get sidetracked like this? Just to be safe, I'll dig out a calculator.

Don't run away from things you have trouble with!

Another thing you might want to get are some 'logic level MOSFETs'
Search this site for discussions about them.
Basically they can be used with the voltage level (5volts) found on the outputs of a 5 volt UNO.

I appreciate the help guys. It was getting ridiculous having to google every little thing I came across. It was becoming a really bad rabbit hole. I needed help breaking the loop :o

thanks for the cool stuff. Ill get started on the vids.

Here is where I am at with my project. Is there a tutorial that will get me a bunch of leds fading in and out?