I have tried to replicate this and it wouldn't work. The switch I have only has two connectors not four like the one in the picture. I wouldn't think that would change anything. Any ideas?

 

Please post a schematic and describe why you think the circuit isn't working.

 

That is the closest I can come to drawing my breadboard since it only gives me pictures of a four post switch and wont let me rotate the components.
My actual switch looks like

and I put one post in F10 and the other in E12.
I put the short leg of my LED in the columm withe the - sign and the long leg in the columm with the + sign.
The resistor with the red wire coming from it in the picture doesn't really have a wire coming from it. The software wouldn't let me rotate the resistor picture. It goes from D11 to E9.

I honestly don't know what is wrong. I have never made a circuit with a transistor before, so I assumed I didn't put the resistors in the same rows as the legs they should actually correspond to.

 

Have a look at how breadboards are wired.
http://www.fiz-ix.com/2016/04/breadboard-wiring-diagram/
You have the LED and one resistor shorted.
Try to draw a circuit first.

 

I know that electricity flows down the - and + columns then extends to the rows when they are connected to the columns. Is the actual picture right? I didn't take that picture. Its from a site I looked at and I followed that.

 

If that's how you have it wired, it won't work.

The collector has no connection to power, the LED is shorted, and there is no power.

 

Oops, sorry.
The resistor is ok. I noticed it was drawn in a shorted position but the red wire indicates the actual connection.
Still, your circuit is wrong. Can you work out how to draw it?

 

Its ok, but no I can't work it out. I just figured power flowed throughout the transistor from the switch. I don't know how to do it.

 

Have a look at...
https://www.dummies.com/programming...ics-projects-how-to-build-a-not-gate-circuit/
That also has an extra LED to help explain it.

 

Is the actual picture right? I didn't take that picture. Its from a site I looked at and I followed that.

No.

1. There's no current limiting resistor for the LED.
2. There's no visible ground connection to the bottom power rail.
3. There are 3 different pin outs for TO-92 transistors. Not knowing the part number, we can't tell from the picture. The only thing we can surmise is that it's not a 2S (Japanese style transistor). BC and 2N have the base in the middle, but the collector and emitter terminals are swapped.

This is how you want to wire:

I changed R2 to 10k; 1k would work, but is overkill. If you increase LED current, base current needs to be one tenth of that.

 

Is it that in the picture the wire is on the same row as the collector and it should be on the same row as the emitter?

 

Unless the breadboard is unusual, there is no connection between one power distribution strip and another (e.g. the "red" strip at the top does not connect to the red strip at the bottom). You have to make the connection with wires.

 

This is the NOT gate circuit but with the extra LED in the base circuit to give a better indication. When one LED is on, the other is "NOT" on.

 

2N3904 is the resistor I am using. Its pins go Emitter, Base, and the last is the collector right? That means the wire in the picture should be coming from the Emitter and not the Collector, correct?

 

The wire from the emitter to batt -ve is ok. But can you see the the resistor from the collector is going nowhere? Also, the LED needs to be between the collector and emitter. Have a look at the circuit I posted above. And, do you realize the LEDs have to be the correct way around?
And LEDs always need current limiting resistors. Do not treat them as light bulbs and place then straight across the power without a series resistor.
In the circuit above, R2 limits the current through LED2, and Q1 shorts LED2 out when turned on so LED2 goes out.

 

If you like, I can wire one up here and post a picture. I just have to get dressed (it is early morning) and get some stuff from my workshop.

 

The wire from the emitter to batt -ve is ok

The wire in that picture is coming from the collector. It would go emitter, base, collector on a 2n3904, meaning the resistor is comming from the emitter and the wire from the collector in the picture, right?

And LEDs always need current limiting resistors.

Yes. I have used an LED before and know how it works. I don't mean that in a disrespectful way and I appreciate your help. I'm just saying its the transistor that I have not used before.

In my mind, the current flowing through the first resistor to the switch had to be enough because I trusted the picture on the site.

However, now that you guys let me know whoever made the picture is wrong I think I see that the wire needs to be in the same row as the emitter, which would be to the left of where it is in the picture, where the resistor going to nothing is, right?

 

Here is my effort.

I use a link as the switch

 

In your first photo in post #1, you failed to show us the complete breadboard.
We cannot see what else you have connected at the bottom end of the breadboard (left hand side of the photo).

Do you know that you have to connect the bus rails on both sides of the breadboard?

 

In your first photo in post #1, you failed to show us the complete breadboard.

That first picture in post 1 is a picture from a tutorial I used. I didn't make that.