Hi there I am curious as to how you would solve this circuit to know voltage drops and currents. It came about from a previous question I had asked which showed my understanding was not quite what I thought it was.
I have looked at other questions of the exact same circuit but again I could not find an answer for myself. I will reference the sources I have found in the bottom. reference 2 seemed to answer my question but I don't understand how to solve the equation, reference 3 had a great explanation but subsequently didn't answer anything for me and reference 1 (and other sources) implied to view it as a voltage divider which would make sense looking at the circuit but when doing my own simulations it didn't seem to add up.
A lot of people talk about using 'thevenin' but I have never came across this before in all my online videos/guides/tutorials/textbooks/highschool physics classes. I am still a beginner and am trying to self teach so maybe I am just not at that level yet.
The LED is a red 2v.
From reference 2 the circuit is the same with different values.'Eugene Sh' from that question states I would find the currents by having the voltage at the node after R1 as V1. Then finding the currents as
I1=I2+I3: (5−V1)/210 = V1/210+(V1−0.6)/200 and they solved it for a V1= X.
My First Question: Is that how you would solve every equation like this? And just use algebra to rearrange that sequence to make it: V1 = XXXXX? If so I dont know how to do that. as I get stuck moving the 2 divisions across. But if that is the way, I can look up algebra lessons or if someone wouldn't mind trying to show me how that would be appreciated.
Reference 1 stated to look at it as a voltage divider
V1= Vin x (R3/R1+R3) Which for my circuit: `V1 = 5 x (21 ÷ (50÷210)) = 0.403V
But that seems way too low (unless I made a mistake). as i think my simulation says it should be 3.68V With a drop of 1.32V across R1.
My Second Question: Do you use this situation as a voltage divider? and if so what did I do wrong.
My Third QuestionDoes it matter if R1 is before or after the ResistorLED-Resistor section? As I thought the total resistance would be the same regardless of R1's position before/after.
My initial understanding was to drop the supply voltage by the LED's forward voltage and use ohms law. However From my previous question asked I learnt that both resistors ARE NOT in parallel. Therefore it is confusing for me with my understanding as I thought that the voltage would split at the node. But now I don't understand how to find the total current with the split not acting as parallel resistors. Originally R1 was after the split creating a 5v divide across each lane. But I changed it to have it uniform with my references. I understand All voltage in must equal negative voltage out. and that current prefers the least resistive path.
I was getting help on my old question but I thought it is more appropriate to create a new question. please let me know if I am posting questions incorrectly as I am still new here.
Thank you for everyone's time.