What I really don't understand is how are both of these not exactly the same once the button is pressed? Closing the circuit on top and bottom produces the exact same circuit to me. One just moves where the "break" in the circuit is, but otherwise are exactly the same once pressed.
Also, how is the top one 5V ( switch off) and 0V (switch on)? I know electricity goes on all circuits (I used to think it would take the path of least resistance) now, so whether it is pressed or not shouldn't it be getting the same?
Sorry about my rookie questions and any help would be appreciated.
Sort of, not exactly. It takes whatever paths are available to it, with less of it taking the path of high resistance and more the path of low resistance. The current will apportion itself according to the resistance there is in the circuit.
You need to understand Ohm's Law and Kirchhoff's circuit laws
These laws are fundamental to all electrical circuits and if you learn them a lot of the mysteries of electricity will become clear. In many cases applying both laws to a circuit will provide whatever answer you are looking for.
Thank you! The bottom one made me re-think the top one as I held it as truly representative of the opposite (0v open, which is obvious.... and 5v closed which confused the heck out of me). So it's a typo.
I looked up the references (thank you for them). I understand that the sum of the inputs at the junction are going to be equal to the outputs. This means that the electricity passing through R2 and S1, in the first image, should sum to that passing through R1. Using the Ohm's law, the electricity going through at R1 is .0005 amps. Right? So I'm going to assume that once S1 is connected all that juice is going to go through that and if any goes through R2 it'll result in negligible amounts. Glad to see others tell me the bottom one is bonkers and does not achieve the explained result in the tutorial I am following.
Where did you get 0.0005A from? R2 is connected to a pin, the current through R2 goes into / out of the pin. Assuming the pin is configured as an input then the input resistance of the pin is in series with R2. I will leave you to work out the rest.