# Why is voltage properly siphoned off in this circuit?

This comes from the Raspberry Pi "fart detector" project. It says to place a 47k Ohm resistor connected to ground in order to siphon off a certain amount of voltage so that the voltage read from the GPIO pin is less.

Given that the row in a breadboard is connected by the same metallic element (thereby electrically common), why doesn't the majority of current flow through to the GPIO pin? Is this because the GPIO pin is acting as voltmeter and thus providing very high resistance? If this logic is true then could the resistor theoretically be placed at J1 and have the same effect?

EDIT: Experimenting with this configuration I found that modifying the arrangement of the resistor relative to the GPIO input didn't make a difference which fundamentally makes sense. When completing everything even moving the resistor ladder around didn't matter as well. The best way I can explain this to myself is that the voltage is present in the same form as long as you are on a common ground.

I wasn't able to find out much about how the input of a GPIO actually works but it seems like it basically draws barely any current and acts as a voltmeter. This allows all 5 resistors to do their job to regulate the current.

(source: raspberrypi.org)

(source: raspberrypi.org)

• Yellow-Purple-Red resistors are 4.7k (shown in diagram) rather than 47k (described in text). Oct 4, 2014 at 19:22
• I am referring to the 47k resistor which is R0. R1-4 should be irrelevant but that was the only schematic I can find. Oct 4, 2014 at 19:25
• I see, I just saw it on the Fritzing diagram and thought I would point it out in case it was causing issues. Oct 4, 2014 at 19:27
• Oh no worries. The Fritzing diagram is actually showing a 47k as well but the orange is hard to distinguish from the red. Oct 4, 2014 at 19:36