Compared to Exp-4a, with the addition of the 470 Ohm resistor, the LED will survive even with the potentiometer all the way close to zero Ohm. I have added the Volmeters to the diagram in order to follow what the book is doing. My understanding is as follows. EXPLANATION: I am assuming that we set the potentiometer at 1K Ohm. Resistance in the circuit = R = 470 + 1,000 = 1.47K ohm. DC power supply = 9V. This implies I = 9/1470 = 6.12 mA. The LED that we have is rated for a 2V drop at 20mA. The implied current in the circuit of around 6.12 mA is far from the rated 20 mA, so the LED will not drop the rated 2V but 1.85V instead. The 1.85V drop with this current instead of the rated 2V/20mA is implied by the EC model and we do not have access to the details of that calculation. It can be estimated via a schokley equation of by looking at the specs of the LED being used. Anyway... Since the LED drops 1.85V from the original 9V, we are left with E=7.15V and R=1470 Ohm (Resistor+Pot. set @1K Ohm). It means that taking into account the impact of the LED, the current in the circuit will be: I = 7.15 / (470+1000) = 4.864mA. The Resistor makes up 470/1470= 31.97% of the total resistance in the circuit. It means that the potential difference of the Resistor is 7.15*(470/1470) = 2.286V. It implies that the potential difference of the potentiometer set at 1K Ohm is 7.15 - 2.286 = 4.864V (also equals 7.15*1000/1470). We match all the results observed in the simulation.
