Fortunately, modern metallized film caps rarely fail with a short, they typically suffer from capacitance loss. It is actually quite hard to tell if disconnected/open PFC capacitor in HX autotransformer circuit can cause somewhat elevated current in some sections of the primary coil.

I tried a simple excersize:

Assume a 70W HX ballast with 240V OCV and two inputs, one 120V and second 240V, the PFC is connected to the 240V input tab, delivering 1A into about 75V load.

Assume common configuration, where the secondary starts from the 120V, so we get 3 windings, each with the same number of turns. I would guess not too far away from a 70W pulse-MH.

This way we have 3 winding sections in total:

I - call it "primary", from N to the center tap (the 120V input), wound on the "primary" side of the core.

II - call it "compensation" from the center to the PFC capacitor/"240V" tap

III - call it "secondary" from the center to the lamp output. This section is on the "secondary" side of the core, so behind the magnetic shunt, so exhibits high leakage inductance towards the rest.

Assume there is no gap in the main magnetic circuit, so the magnetizing current with no load is zero (a bit of simplification).

The secondary sees the full lamp current, so the 1A, in any case. Assume the lamp is good and on spec.

First the scenario with no PFC connected and feed from the 120V input (is first, because it is the simplest :-) ):

Because the secondary imposes 1A times the number of turns on the core and there is just the primary winding connected, the same 1A has to flow through the primary section too to cancel out the magnetizing current (no gap in the main magnetic circuit). So the mains current is then 2A.

So in this scenario we have

Primary with 1A

Compensation with none

Secondary with 1A

The ballast is not primarily designed to work this way.

Second scenario is fed from 240V input. Does not matter if PFC is there or not, as it is connected straight to mains.

Here the secondary field has to split among "primary" and "compensation" (here acting as a part of primary). Because the way how the windings are connected, the "primary" sees a difference between "secondary" and "compensation" winding currents.

If you put the equations together, you get

Primary with none,

Compensation with 1A,

Secondary with 1A

The ballast is designed to work that way.

The 3rd scenario is the most complex one, feeding 120V but the PFC capacitor present (assume set for PF=1 for simplicity).

Here the total 1A is handled by the "secondary", but then gets split to "primary" and "compensation".

Assuming the real total power input (so lamp power plus ballast losses) is 100W, so the mains current will be 0.83A.

The apparent power without compensation is OCV * 1A = 240VA. Half of this is handled by the "secondary", the other half needs to split among "compensation" and "primary".

So the reactive power for the PFC to handle is 218VAr, so 0.91A at the compensation winding.

The "primary" will then handle half of the real power, so about 0.416A.

So summary:

Primary has 0.415A

Compensation 0.91A

Secondary 1A

The ballast is designed to work this way.

So you can see the both ways the ballast is designed for, the secondary and compensation winding currents are about 1A, so both windings will be designed to handle that for most of the time. But the primary sees either no current, or just 0.41A.

But when fed from 120V with no PFC, the primary will see 1A instead of the 0.41A it is normally supposed to see, so technically about 2.5x current overload.

But what matters is not that much the currents, but the total power dissipation in the ballast, as that is the main factor determining the operating temperature. The power dissipation is I^2 * R for a given winding.

Now to minimize the total dissipation in a transformer, the amount of copper for each winding should follow the same ratios as is the VA each winding has to handle. And the winding resistance is inverse proportional to that cross section. Because we have all windings at 120V, ratios of currents will be the same as the VAs, assuming the manufacturer really optimizes to the last T and choses exactly those wire cross sections.

As the 120V input (with PFC as without is not the design target) is the worst case, we take:

So primary will have relative resistance 1/0.41 = 2.55,

Secondary 1/1 = 1

Compensation 1/0.91 = 1.1

So normal relative power dissipation will be 0.42^2 * 1/0.41 + 1^2 * 1/1 + 0.91^2 * 1/0.91 = 2.33

Same for the 240V input: 0^2 * 1/0.41 + 1^2 * 1/1 + 1^2 * 1/0.91 = 2.1

Now when removing the PFC: 1^2 * 1/0.41 + 1^2 * 1/1 = 3.44

So the power dissipation (so the temperature rise) will be about 1.47x higher than with the PFC. Comparing to plain higher current in all windings, it is equivalent to about 1.25x higher currents.