The LED forward voltage drop will remain (roughly) the same, but the current can change, so the calculation becomes (same equation solving for I):
$$I_{LED} = {(V_s - V_f)\over{R}}$$
So for a 3V \${V_f}\$ and a 5V supply, the \$100\Omega\$ resistor would give \${(5V - 3V)\over{100 \Omega }} = 20 mA\$.
So if you know what current you want, just plug the values in, e.g. for 10mA:
$$R = {(5V - 3V)\over{0.01 A}} = {200 \Omega}$$
Basically, the fact that the supply and the LED forward voltage can be relied upon to be pretty static, means that whatever value resistor you put in will also have a static voltage across it (e.g ~2V in this case), so it just leaves you to find out that voltage and select a resistance value according to the current you want.
Below is the V-I curve of a diode (from the wiki LED page), notice the current sharply rises (exponentially) but voltage stays roughly the same when the "on" voltage is reached.
For more accurate control of the current you would use a constant current, which is what most LED driver ICs provide.
it appears to me that the current generated by 35 V and 2vx will
collide each other
It may be that you are assuming that a voltage source, whether independent or controlled, must source current, i.e., supply power to the circuit.
But, at least in ideal circuit theory, there's nothing "wrong" with a voltage source sinking current, i.e., receiving power from the circuit.
For a real world example, consider that, when a battery is being charged, the current is in the opposite direction than when the battery is being discharged.
I would like to know how the current flows across 5 Ω resistor.
If you're planning to be an EE, don't write or say things like "current across"; current is through, voltage is across.
Now, this circuit is very easy to solve. There are two unknowns so you need two independent equations.
For the 1st, write a KVL equation clockwise 'round the loop:
$$35V = v_x + 2v_x - v_o \rightarrow 3v_x = 35V + v_o$$
Now, you need one more independent equation. Can you find one?
Best Answer
There is one important low-impedance node in the whole circuit. That's the frame and the cross in the middle. Let's call that 0v. The top voltage source is connected to the middle via that 1k resistor, that means the lower voltage source has priority.
The current flowing in: