I am using the AoE lab book to teach myself electronics. I am on the voltage divider section, and I am confused at their simplified explanation. I'd like to explain it a different way, but I am not sure how to understand the open circuit voltage.
For the following circuit:
simulate this circuit – Schematic created using CircuitLab
Calculating Vout is simple:
$$V_{out} = V_{in}\frac{R2}{R1 + R2}$$ $$V_{out} = 30\frac{10000}{20000} = 15V$$
Now we introduce a load into the circuit:
My approach to solving this myself was to turn the original circuit into a black box and imagine we are looking at it from the perspective of the output.
From reading, the short circuit current here will be
$$I_{sc} = \frac{V}{R} = \frac{15}{10000} = .0015A$$
The first of my questions is - why does only the first resistor matter for calculating the short circuit voltage? Shouldn't both resistors be...resisting in the circuit? Intuitively, is this because we are bypassing R2 via a short circuit and the voltage will just (for lack of a better term) ignore the resistor and follow the path of least resistance (ground)?
After this I need to derive the open circuit voltage. I am having trouble figuring out how to view the circuit in a way that makes this make sense. If I disconnect the load, the open circuit voltage would be the original Vout, would it not?
Basically, I am trying to derive the math that makes treating R1 and R2 in parallel make sense. I understand the thevenin resistance will be this value, and the thevenin voltage is equal to vout (15V) in this case. The explanations in the lab book, while helpful, sort of handwaved through this part (they began to talk about switching voltages and grounds to show which resistor matters) and I'd appreciate any help in understanding this section. I feel like it is critical to understanding circuit impedance (the next section) and I would like to fully understand it before moving on.
Thank you!