It's been a few years since I first studied the Ohm's law at school. Now I'm really getting into electronics, and I must admit a part of it still puzzles me.
I know this is kind of a strange question, but I have no clue how to express it otherwise, so I'll just lay out my reasoning. The following train of thoughts is wrong and/or incomplete. Please stop me whenever I'm wrong (even for the sligthest thing), and complete this reasoning.
Here it comes:
Let's make a circuit with:
- 6V battery (typically 4 AA batteries)
- 2k ohms resistor
- LED, for the sake of doing something
Circuit map:
simulate this circuit – Schematic created using CircuitLab
From Ohm's law, we know that the current in the circuit will be of (considering only resistor has any resistance)
\$ I = \dfrac{V}{R} = 3mA \$
I do understand that the current is limited due to the resistor... crystal clear.
The book tells me that, if I measure the potential difference between both sides of the resistor, I will have something between 0 and 6 volts.
Sounds legit:
- can't be 6 volts because the two ends of the battery are connected, therefore lowering the potential
- can't be 0 volts, since the current does not freely move around, due to the resistor; resulting in one side being more negatively charged, and the other more positively charged, since the electrons can move faster to an end of the battery than through the resistor.
What I don't understand (and if my whole reasoning is right until now), is: How can I calculate the expected voltage around the resistor ?