In a circuit with a resistor parallel to a diode/resistor, will the voltage across both resistors be the supply minus the 2v drop of the LED?
Meaning you would calculate it still by using resistors in parallel: 1 ÷( 1/r1 + 1/r2). And then use supply minus 2v ÷ R to get the total current? Is this correct?
I am trying to teach myself electronics and find it very hard to find definitive answers on the things I am trying to learn. Therefore I am wanting to just try get confirmation on a question if I can.
I get confused as in a parallel split the voltage is unchanged on each wire, only current is halved. But since one lane has a voltage drop I initially assumed they would rejoin with different voltages which wouldn't make sense. And I have learnt that if there is 2 diodes it would only activate the one with the lowest forward biasing voltage and the other would stay off. .
I have done a few tests and found that seem to imply this is the case but none seem to definitively say it. If someone would be and to assist in confirming that this is the case that would be appreciated.
I have been learning about diodes and LEDs but keep running into problems that tutorials haven't discussed so it's tricky to find all the information. Apologies if I'm not posting correctly this is my first post. Also I know there is a circuit editor but I can't seem to see it in mobile. Thank you
EDIT: Schematic for a comment.