Perhaps a worked example would make things clearer.
Suppose I had three lamps - each has a different working voltage and takes a different current. What would happen if I connected all three in series and connected them to a 9V battery.
For the purposes of illustration only (keeping it simple) I am assuming the lamps work as pure resistances that don't change their values.
First we calculate the individual resistances of the lamps. In this case 6,24 and 15 ohms.
The total circuit resistance will be 6 + 24 + 15 = 45 ohms
The current that will flow around the circuit will be
9/45 = 0.2 Amps
This is much less than any of the rated values because there is MORE resistance in the circuit
Each lamp will 'drop' a different voltage
lamp1 will drop 0.2 x 6 = 1.2V
lamp2 will drop 0.2 x 24 = 4.8V
lamp3 will drop 0.2 x 15 = 3.0V
If we add up all the voltage drops we get 1.2 + 4.8 + 3.0 = 9V
In other words the sum of the voltage drops around the circuit is equal to the supply voltage (kirchoff's voltage rule)
Again note that these voltages are not related to the ratings of the devices.
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
If the sources were ideal it would be solvable by inspection:
Since all the components are in series, the current around the loop is 5 A.
Point C is at ground.
Point B is at +20 V.
Point A is at +520 V.
The unlabelled node between the current source and V1 is at 500 V.
Since you have said that the current source is only capable of 10 V, it is overloaded. Possibly its output voltage sags to 10 V and the current is reduced to 100 mA. Possibly its protection circuit operates and it becomes an open circuit or a short circuit. Possibly it catches fire and burns your house down. Without a clarification on what the voltage rating means, the actual result can't be determined.