Equations for a simplified transistor circuit
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. CircuitLab version allows simulation.

This is just a rough estimate approach.

SW1 open


With R1 = 10k we'll get 0.1 mA per volt difference between V1 and Q1's base. If the LED turns on it will clamp Q1's emitter at around 2 V. That means that Q1's base will be in the region of 2.5 V. IR1 will be about 0.95 mA. Let's say 1 mA.
If Q1 has a gain of 100 it should be able to pass a 100 mA collector current.
It can't. R3 is 680 &Omega; so the current will be limited to \$ \frac {12 - V_{LED}}{680}= 15 \ \text {mA} \$. This is a safe value for an indicator LED.


SW1 closed


With SW1 closed R1 and R2 form a potential divider of 11:1. The maximum voltage that will appear on the base is \$ \frac {12}{11} = 1.1 \ \text {V} \$.
The emitter will be one diode drop lower so about 0.4 V.
This isn't enough to light the LED. It will be off.


You can set this up in CircuitLab, click on Simuulate | DC solver | probe the top of the LED which will add a voltage and current measure to the measurement list | Run DC Solver. The results agree pretty well with our very rough calculations.




  (and simplify it so it acts like an on/off switch).


This part of your question is vague.







simulate this circuit

Figure 2. (a) The gross simplification. (b) A modest simplification.

Figure 2b might be useful to control the LED via a really sensitive switch that couldn't handle the 15 mA LED current.