LEDs are P-N semiconductor devices. The current follows the voltage in an exponential equation. Once the applied forward voltage exceeds a certain minimum voltage, (turn-on, knee, forward), the current takes off exponentially. Because of this behaviour the diode acts like a crude voltage regulator holding the voltage at the forward voltage.

You don't worry about the resistance of the diode simply because it is not constant. Get a rough estimate of the forward voltage, Vf as a 1st approximation.

Take the source voltage Vs and subtract Vf.
(Vs - Vf) is the voltage across the series resistor.
Hence the current through the resistor (and the diode) is Id = (Vs - Vf)/ Rs

At this point we have ignored the current through the other diode in parallel with the first diode.
There are only two ways to calculate the current through the second diode (not including actual measurement). You know the voltage across the diode which is the same across the first diode.

1) If you know the I-V equation of the second diode then you can calculate the current Id for a given voltage Vd.

2) If you have the I-V curve of the second diode then you can estimate the current Id by looking it up on the chart. This is the same as (1) but in a graphical form.




Read about diode load line.



Note that in either case, the I-V characteristic is also a function of temperature. The self-heating effect of the LED will contributed to this effect.

After you have done the above, you can include this current in a 2nd iteration and see how much it alters the 1st approximation.

 

Thank you very much for all you answers, it has been a long time since I found a community where I got so much help. Give me some time and I will beginners as well.

Thanks @MrAI but your math overwhelmed me, because this circuit was shown on this channel so early (before he explained how to calculate resistance), I thought the math would be on similar level. I am also not sure how to work with such number: 4.6128070020549152e-4 it must be 0,00046 if I am not wrong.
I also understood everything your wrote and replaced correctly all the variables with x,y,z and entered them in few online equations solver, with no luck.

All the answers merged to the following thoughts:
Only one LED is emitting visible light, because it drains all of the 23mA ( I = U / R → 9V / 3.9k Ω = 0,0023 A) or
18mA ( @MrChips I = (Vs - Vf) / Rs; (9V-1.8V)/3.9k Ω) not sure what we get here, is it the correct current, because in the calculation above I didn't added the resistance of the LED?
Higher Vf, more resistance and smaller current. Red LED has smaller resistance and all the current flows though red.

What about voltage drop? Because the LED needs the 1.8V, we have a Vdrop of 7.2V at the resistor? But we have a branch, we should have same voltage on another branch.
2x1.8V=3.6V-9V we get 5.4 voltage drop at resistor.
If we use Ohm's law
3.9k*0.002A=7.8V
3.9k**0.0023A=8.97V
3.9k*0.0018A=7.02V (I = (Vs - Vf) / Rs)....... 9-7.02-1.8=0.18V left for blue
this value looks a lot better, because I think there is no current flowing to another LED, because we don't have 0.7V to overcome barrier potential of the second LED. But I am wrong, MrAI calculated a current of 0.5mA but he used a much smaller resistor... so his version of the circuit the blue led could has more than 0.7V. How do I even calculate the voltage on this branch? U=R*I, 390 Ω*0.0005A= 0.195V. Why is there a current flowing when we have not enough voltage to overcome the depletion layer?

I am sorry but I don't know what happens to the voltage in this circuit. Because red is glowing, it has a Vdrop of at least 1.8V, which voltages have the other LED?
The asked question was why red glows, I learned because it has smaller resistance, based on the smallest LED Vf value. But now I am not sure what to think about the voltage behavior in this circuit.

Thanks again guys. I am also available for a private lesson on any VOIP software just in case... I live in Germany UTC+1

Here is another more intuitive solution. It involves varying just one parameter vLED with a separate voltage source until we see the circuit currents solved. See the two diagrams for the schematic of the test circuit and the plot we get later in this procedure. This is a very interesting way to solve this and you can even do it in real life if you are careful with the adjustment of the second voltage source, and the second voltage source must be able to source and sink current to work.

The circuit is shown below. We have a 9v constant voltage source V1
and a 0 to 3v variable voltage source V2.
We have resistor R1 which is the series resistor, and a second resistor
R2 which we can use for this experiment. Although R2 is shown as 1 Ohm
that is very low so it has little effect here so if you like you can
think of it as zero Ohms. WE use that resistor to sense the current
through the two LEDs independently from the current through R1 as we
vary the V2 voltage.

By looking at the diagram, we can see that if we vary V2 and measure iR1
and iR2 we will eventually reach a point where iR2=iR2 and that would mean
that we have the required voltage that solves the question of what the
voltage across the two LEDs is (remember we can think of R2 as being zero
Ohms here). So the idea now is to vary V2 and wait until we reach the
point where iR1=Ir2.

We can do this experimentally being careful to start V2 at 0v and raise
it very slowly as we measure the two currents, or we can do it using meth.
To use math we need the two LED equations for current.

So we start with these two LED equations:
Id2=(29294687*(e^(2.0*Vd)-1))/78507051450
Id3=(3603849*(e^(2.0*Vd)-1))/72514630250

where Id2 is the current though LED2 which is a 2v (at 20ma) nominatl voltage LED,
and ID3 is the current through LED3 which is a 3v (at 20ma) nominal voltage LED.
Vd is the as of yet unspecified voltage we want to solve for, and it will be the
same for both LEDs because they are in parallel.

Since LED3 has a higher nominal voltage at the same nominal current we would
expect the current through LED3 to be lower than LED2 but we can wait to see
the result after we do a few simple calculations.

Now we can see that the two LEDs are in parallel, so their currents add and the
total sum flows though R1 if the second voltage source was not present.
Thus we add the two currents Id2 and Id3 and we get:
Isum=(24072209549555128*(e^(2.0*Vd)-1))/56929098079144763625

and we can also calculate the voltage at the right side of R1 by noting that
if we knew the sum of currents (as above) we could multiply by R1 to get the
voltage drop in R1 and subtract that from the 9v source, and that gives us
the two LED voltages as they are in parallel (again think of R2 as 0 Ohms).
The votlage drop across R1 then is:
vR1=R1*Isum=500*Isum=500*((24072209549555128*(e^(2.0*Vd)-1))/56929098079144763625)

Now we can subtract that from the 9v source to get the two LED voltages:
vLEDs=9-500*((24072209549555128*(e^(2.0*Vd)-1))/56929098079144763625)

and now we end up with an equation that has the LED voltages on the left and
the voltage on the right side of R1.

Now a little logic helps here next. Since we consider R2 to be zero that means
that the voltage on the right of R1 is the same as the two LED voltages. So
we can set vLEDs=Vd as well. Doing this we get;
Vd=9-500*((24072209549555128*(e^(2.0*Vd)-1))/56929098079144763625)

This is where we vary Vd which in the circuit is also V2 (again R2=0).
We vary Vd and measure iR1 and iR2 and when they are the same that means
we have the right LED voltage setting of V2.

To do this mathematically, all we have to do is solve that last equation for Vd
numerically. This is easy too because there is only one variable.
Since we ended up with this:
Vd=9-500*((24072209549555128*(e^(2.0*Vd)-1))/56929098079144763625)

we can note that when Vd is the right set value that means that both
sides of that equation will be equal, and that in turn means that
if we subtract the right side from the left side we get zero:
Vd-(=9-500*((24072209549555128*(e^(2.0*Vd)-1))/56929098079144763625))=0

Now to solve this graphically all we have to do is take the left side of that:
Vd-(9-500*((24072209549555128*(e^(2.0*Vd)-1))/56929098079144763625))

and plot that while we run Vd from say 0 to 3 volts.
Doing that we get the second diagram which is this graph.
I drew a red line that shows that the zero point coincides with about 1.75 volts.
This means the LED voltages are both near 1.75 volts.

Now as mentioned before we can also solve that equation numerically. Doing this
we end up with a more precise value:
vLEDs=1.779818726269 volts

Two get the two LED currents you can just substitute this value (or the simpler one) into
the two LED equations and you will see the two different currents. The two currents will
be very different from each other because of the way diode curves work. We could
repeat this experiment using linear devices (which would be plain resistors) and see the
difference.

Now this result comes about by using some LEDs that are really quite fictitious.
In real life we will get different results which depend highly one what the diode
functions are.

 

Hello again,

Another interesting circuit to work out is a circuit as before except we place a second resistor in series with the lower voltage LED (that would be the red one in the original circuit) and calculate the resistance needed to force both LEDs to conduct the same current level.

I did this with a 2v at 20ma LED and a 3v at 20ma LED and R1=500 Ohms and found that the resistance needed in series with the 2v LED was about 165 Ohms. That caused the current in both LEDs to be about 6ma each (total current from the 9v source is about 12ma).

It is not as hard as it might seem at first because we have to have the two LED currents the same, so we can set iLED2=iLED1 or iLED1=iLED2 and go from there. That means only ONE current to solve for which amazingly reduces the circuit complexity down to just one LED current to solve for.

If you have never done this it is an interesting circuit math exercise. I might post a solution to one particular circuit in the future.
In real life though the LED characteristics are hard to predict exactly so we look for an approximate solution, or we use a resistance substitution box or a potentiometer to vary the resistance of the resistor connected to the lower voltage LED while measuring the current in both LEDs. When the current is equal in both we have the right resistance value.

Good luck