Unless I'm missing something subtle in your question then....
Primary impedances Rp+jwLp are irrelevant for a current transformer - the current flows and that's it and if those leakage impedances are too high then it's a really badly designed CT.
Magnetization reactance is also irrelevant if the burden is chosen to be non-excessive i.e. the burden shunts the magnetization inductance so that it can be ignored.
Just do some simple math on typical impedances and you should conclude what I've said above.
with a 1 Amp slow blow fuse on the primary
Given that your secondary fuse rating implies a load of 120 VA and your primary fuse rating implies an input of 240 VA it could be argued that the magnetization current into the primary is in the order of 0.5 amps\$^1\$ leaving 0.5 amps for real load current seen at the primary (0.5 amps x 240 volts = 120 VA).
This implies to me that your output fuse should be significantly smaller than 1 amp i.e. more like 0.5 amps because if you were to take 1 amp from the output winding of the dual-connected transformer, you would inevitably be taking 1.5 amps into the primary of the transformer connected to raw mains and this could cause a fire.
is it recommended to install a current limiting circuit (or component)
to limit the output current to below 1 Amp or is this overkill?
Given the info you have provided I have to be cautious and say that the output fuse should be no more than 0.5 amps. You will still need to use a 1 A input fuse too.....
But, there is a further consideration. The transformer that is receiving the main AC input is not only taking magnetization current for its own core but supplying magnetization current for the other transformer's core.
It logically follows that in the dual connected scenario, the 1st transformer might be taking up to 1 amp just to magnetize both cores leaving you with no headroom for load current.
There is insufficient information to safely conclude anything else so, I would recommend you measure the primary current (nothing on the secondary) and, if you find it to be (say) 0.25 amps you can also assume that it will be 0.5 amps when both transformers are cascaded. This would mean that your output fuse should be 0.5 amps.
\$^1\$ This is a bit of an approximation and underlies the basic problem of cascading transformers. If the full rated primary current is 1 amp and this current contains 0.5 amps of contribution from a 120 VA load, magnetization current could be as high as \$\sqrt{1^2 - 0.5^2}\$ = 0.866 amps RMS. The square root calculation takes into account that magnetization current is due to primary inductance and this will normally be 90 degrees shifted to resistive load current. However with two cascaded transformers the total magnetization current could, theoretically be as high as 2 x 0.866 amps. It's very problematic cascading transformers like this.
Best Answer
That is because it is an inductor, not a transformer. The idea is to limit the current by introduce inductive impedance. The impedance is adjusted by adjusting the air-gap.
Either will have an effect. You will need to do more research to get the design details and determine the best scheme for the specific design.
I suppose "rheostat transformer" is what you are calling the variable inductor. The safe level of current depends on the size of wire, the insulation temperature rating, the core dimensions and material, the number of winding layers, external cooling, and whatever I may not have thought about. You probably need to search for variable-reluctance inductor design information.