If we have this circuit:
The diode with the lowest forward voltage drop will conduct and turn on while the other will turn off and not conduct current.
If I have this circuit:
Assuming the 2 diodes have the same voltage drop, R1 can be considered to be in parallel to R2 and that's how we solve those circuits.
However in the same case, what if the 2 diodes have different forward voltage drops? Say Vd1>Vd2 which assumption should I take into account?
You'd need to solve the circuit, considering that the current must split, and the resistor voltage drop (V = IR) plus the resistor diode drop match. You may need to use a system of equations to relate all of the things happening simultaneously, such as the following:
Let \$V_x\$ be the voltage at the green probe, \$i_1\$ the current through D1/R2 (top branch), and \$i_2\$ through D2/R1 (bottom branch). I'll keep your assumption that the voltage drop of a forward-biased diode is constant. and furthermore, I will assume that the diodes are all forward-biased (and will check this assumption after solving).
We can relate the voltage drop between the 5V supply and Vx with the currents in the diode-resistor branches, and we can relate the voltage drop across R3 with the current in that branch (i.e. the sum of \$i_1\$ and \$i_2\$).
This gives us:
$$ \begin{align} V_x &= (i_1 + i_2)R_3\\ 5 - V_x &= i_1R_2 + V_{D1}\\ 5 - V_x &= i_2R_1 + V_{D2} \end{align} $$
This is a system in three equations and three unknowns, and can be solved using your choice of solution technique.
Once you solve it, you must verify that the diodes are, in fact, forward biased (i.e. \$5 - Vx\$ is greater than \$V_{D1}\$ and \$V_{D2}\$). If the assumption is violated, then that diode is actually not forward-biased, and no current flows in that branch; you'll set up the equations and solve them again under that revised assumption.
On the other hand, the constant-drop model is no good in the first schematic you have. The two diodes are in parallel, so their current must split according to the smooth I/V curves associated with each of them. The series resistors added to both branches in the second schematic make the approximation hold significantly better (but still not that well).
The diode with the lowest forward voltage drop will conduct and turn on while the other will turn off and not conduct current.
No, one diode will likely conduct more current than the other, but they will both conduct current.
Assuming the 2 diodes have the same voltage drop, R1 can be considered to be in parallel to R2
No, if the two diodes have the same voltage drop, the resistors will also have the same voltage drop. However, the fact that the resistors have the same voltage drop does not make them "in parallel".
However in the same case, what if the 2 diodes have different forward voltage drops? Say Vd1>Vd2 which assumption should I take into account?
If diodes had a fixed voltage drop (they don't - that is only an approximation) then, for each path, you would subtract that voltage drop from the total voltage across the diodes and resistors, and use Ohm's law to calculate the current in each path.
In reality, the current through a forward biased diode is nearly exactly exponentially related to the voltage across the diode. One can use a computer to solve for the current through, and the voltage across, each diode in the circuit. However, the solution technique generally involves finding successive approximations, and so is difficult for a human to calculate a precise answer by hand.
Addendum:
Why One Cannot Treat Resistors R1 and R2 in the Second Diagram as Parallel.
Suppose that
If R1 and R2 were parallel, they could be replaced with a single resistor \$R_p\$ with a value of $$R_p=\frac{R_1R_2}{R_1+R_2} = \frac{100,000}{1,100} = 90.91\Omega$$
and the circuit would look like
simulate this circuit – Schematic created using CircuitLab
With this arrangement, the current in each diode D1, and D2 would be identical.
However, with the actual circuit, again, assuming each diode is identical and has a voltage drop of 0.7V, we can replace the diodes with voltage sources, like this:
With this arrangement, 2.048mA flows through D1 and 20.48mA flows through D2.
Conclusion, R1 and R2 cannot be replaced by a parallel equivalent resistor, because it changes the behavior of the circuit.
Assuming the 2 diodes have the same voltage drop, R1 can be considered to be in parallel to R2 and that's how we solve those circuits.
Can't do that. That assumption comes from a "constant forward voltage" model of the diode, which is too much of a simplification to be useful when you have multiple in parallel.
The diode with the lowest forward voltage drop will conduct and turn on while the other will turn off and not conduct current.
Also, same model of a diode either not conducting or conducting and having a constant forward voltage drop.
That model is OK if you only want to model rectification, but it's totally insufficient for modelling circuits where multiple diodes are in parallel. Both will conduct.
You need to take the usual exponential current-voltage relation, at the very least, to model these circuits.
