Question About Analog Input and Ground

I apologize if this is the wrong location, but I didn’t see any other place I thought would be better to ask.

So, I am playing around with an Arduino Uno to learn more about electronics, but there I can’t wrap my head around something.

Question start: When a component is connected to both an Analog Input and the Ground (in my case, I have a potentiometer connected to both power and ground, and connected to an Analog Input in order to read a value), why does electricity travel to the Analog Input, allowing me to receive information, instead of just to the Ground?

The pot's wiper splits the resistance of the pot into 2 resistors, creating a voltage divider.

The analog input is really just "sensing" voltage. Very-very little current flows into the analog input. And of course, that voltage is there whether the Arduino (or a multimeter) is connected or not.

The potentiometer is a variable [u]voltage divider[/u].

For example, with the pot at mid-position you have two equal-value resistors and the voltage is divided in half. With a multimeter, you can measure the voltage across the top resistor (or the top half of the pot) or the bottom resistor. But, the Arduino's analog input is referenced to ground so you can only measure the voltage across the bottom resistor (or you could measure the 5V at the top, referenced to ground).

If you take an electronics class, the 1st thing you learn is [u]Ohm's Law[/u] which describes the relationship between voltage, resistance, and current. i.e. More voltage = more current. More resistance = less current. (Resistance is the resistance to current flow.)

At some point, you learn [u]Kirchhoff's Laws[/u] which describe how voltages divide in series circuits and how currents sum in parallel circuits.

As long as you realize that resistances add in series and that the same current flows through both series resistors, you can simply apply Ohm's Law to calculate the current, and use Ohm's Law again to calculate the voltage drop across each resistor in a voltage divider.

It's important that very little current flows into the Arduino (or whatever else is connected), otherwise the current won't be the same through both resistors (or both halves of the pot) and the calculations are not quite as simple.

Go back to the water flow / pressure analogy...

You have a long thin water pipe connected to a header tank, with the bottom end draining out, and a pressure meter you can tap into the pipe where you like.

At the start of the pipe the water pressure is 5 units, lets say, and at the outlet its 0 units.

Half way along you tap into the pipe and measure 2.5 units of pressure - the pipe is resisting the flow and causing a pressure difference (or a gradient of pressure along the pipe).

An analog pin is like that pressure meter - it takes no flow but senses the (electrical) pressure (which is called potential or voltage). All the electrical current flows to ground.

The potentiometer basically moves the tap point along the resistance track to any point, and unlike water we don't have to worry about leaks as electricity doesn't leak through insulators like air (in the water analogy we have to keep blocking off the holes we made for the meter!).

To be exact there is a slight leakage current to the analog pin, but its something like 0.000000001% of the main current path so can be ignored for practical purposes.

Thank you all so much! It makes a lot more sense now, and I have a few directions on what to research to continue learning about it. You guys are awesome! The water analogy really helped me to understand the whole concept of voltage better as well.

Just don't take the water analogy too far...

Yes, keep your Arduino dry!