... Re-evaluation .....
If you are to find the Norton equivalent, then maybe the load terminals are the two lines drawn at the right of the circuit .... with the load not actually given.
Therefore, for Norton equivalence, you find Rn by looking into the load terminal, and 'open circuit' all the current sources. There are two in this case. Then find the resulting admittance/impedance of this circuit. This will require a little complex number manipulation, but should be simple with a calculator.
Next, short the output terminals with a straight line .... to find the short circuit current, looking into the circuit from the load viewpoint. Doing this will actually short out the VAdc current source, and you can find the resulting total admittance ... just the sum of the leftmost resistor admittance, inductor admittance, and then add the center capacitive admittance. Take the reciprocal of this summed admittance quantity, obtaining a total impedance, and multiply by the leftmost current source. This resulting quantity is the complex voltage at V. The one remaining step is to take this complex voltage at V quantity, and divide it by the impedance of the center capacitor... or multiply by the center capacitor admittance. By doing this, you will find the short circuit, or Norton current.... That is, the current that will flow through a short circuit across the load terminals, and also the center capacitor element.
It seems that it is very convenient to use the Norton equivalent, in that the variable current source, VAdc, is shorted out during the calculation of the Norton short circuit current at the load terminals.
You have calculated the short circuit current ... at the load terminals .... the same as that in the j2ms center capacitor.
You have found the Norton Resistance, Rn.
Therefore you have found the Norton equivalent of the original circuit.
Does this seem to make sense? Surely, the underlying principles were provided in your reference materials....
I get:
Rn = (49.1-j4.97) Ω .... or in terms of admittance: (20.2+j1.94) ms
In= (0.587+j3.91) mA
.... will try to check this result later.