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.
After step 6, your circuit will be as shown below.
simulate this circuit – Schematic created using CircuitLab
Which can be further reduced as a 16V battery in series with 3k and 5k resistors.
Now, you will have 10V across this 5k resistance, ie, 10V across each parallel 10k. Splitting 10k into 6k+4k, you will have 4V across 4k resistance. From the polarity of v1 marked, V1= -4V.
Best Answer
I've worked in electronics all my life (and am now retired). I perhaps did do calculations based in Kirchoff's laws, current division, etc in school, but now just consider Kirchoff's laws to be scientific wording of what should be simple, common-sense observations.
When I look at this problem, I see the 0.1 mA through the 60K resistor. I see that there is a resistor of twice that value (120K), in parallel, so it will have half the current of the 60K, so the total current in the circuit will be 0.15 mA.
The voltage across the 60K resistor is 60K x 0.1 mA, or 6 volts.
The voltage across the 20K resistor is 20K x 0.15 mA, or 3 volts, so Vs is 6 + 3 = 9 volts.