If you wish to solve the circuit using node voltage analysis, you would not bother to write a KCL equation at node 2.
Remember, when doing node voltage analysis, one is solving for the node voltages.
But, the voltage at node 2 is given: \$V_2 = V_{ab}\$
So, you might think that you must write a KCL equation for node 1 but, in fact, you don't because there is a voltage source connected there too.
Simply use KVL to write:
$$V_x + 0.01V_x = V_{ab} \rightarrow V_x = \dfrac{V_{ab}}{1.01}$$
Now, you know the node voltages so you can find the resistor currents. Can you take it from here to find \$I_a\$?
Finally, about Ia. I am also confused by the presence of 0.01Vx. Would
applying KCL only means finding current between node 1 and 2 or do we
have to involve 0.01Vx too?
Since you know the node voltages, you know the currents through the resistors connected to node 1. Thus, if you write a KCL equation there, the only unknown is the current through the dependent source so use this KCL equation to solve for the dependent source current.
Now that you've found the dependent source current, KCL at node 2 involves only one unknown current, the current \$I_a\$.
The reason why I applied node analysis is because I am studying it
these days, and wanted to apply it correctly. Did I?
To correctly apply node voltage analysis, you must enclose the dependent voltage source and parallel resistor inside a supernode. The KCL equation for the supernode is:
$$\dfrac{V_x - V_s}{25} + \dfrac{V_x}{150} = I_a $$
There are two unknowns so you need another equation which is the KVL equation I wrote above.
Note that the 50 ohm resistor is not a factor in the equation. This is due to the fact that it is in parallel with a voltage source which means that the only circuit variable the 50 ohm resistor affects is the current through the dependent source.
When analyzing a circuit, you can put the arrows in either direction according to whim, a flipped coin, or Tarot cards.
After applying Kirchoff's laws to compute all the voltages and currents, you'll find some variables have negative values. Those correspond to arrows you drew backwards. Fix those, and then you know the directions of currents in all branches of the circuit.
It is perfectly normal for an experienced engineer to get a few initially backwards, when multiple different voltage sources are pushing in opposite directions. You can only guess, and let algebra tell you the net result.
Best Answer
Question statement is a little hazy, but the general idea is, current entering a node gets a positive sign, current leaving a node gets a negative sign, and the sum of all of them comes to zero.
One way to assign currents is to recognize that current flows around the loops. So for each loop, you name a variable for the current. Then it should be straightforward to enumerate the currents entering and leaving a node.