I did some calculations of the resistance of various light bulbs based on their wattage ratings, which reflects their "steady state" resistance following the brief warmup that occurs at turn-on. I also measured their resistances with a good multimeter, which reflects the "cold" resistance they would very briefly have just after turn-on. The bottom line in each case was well over a 10:1 difference between those two resistances.
Which makes using a light bulb a tricky matter at best. Depending on the fuse rating and the wattage of the light bulb you would probably have either too little "steady state" current to be effective in a reasonable amount of time, or too much current during the instants following turn-on, which might blow the fuse.
I also thought about the possibility of using a resistor instead of a light bulb, since a suitably chosen resistor would provide essentially the same resistance immediately following turn-on as it would subsequently. However the problem with that approach, assuming you would be powering the setup with 120 volts AC, is that the power handling capability of the resistor would have to be very large.
So what I would suggest is that you use a Variac in conjunction with the light bulb approach. I suspect that you can find a used or possibly a new Variac inexpensively at eBay. That would allow you to bring up the AC voltage applied to the light bulb/fuse combination slowly, to avoid excessive current flow while the bulb is heating up.
The resistance of the fuse itself would be negligible compared to the resistance of the bulb. The resistance of the bulb can be calculated from its wattage rating (its actual wattage rating, not the "incandescent equivalent" rating) based on the following relation:
R (in ohms) = (120 volts squared) / (wattage rating)
So for example a 53 watt halogen bulb having a 75 watt incandescent equivalent rating would have a steady state resistance of:
R = (120 x 120)/53 = 272 ohms.
The resulting current would be:
I = E/R = 120/272 = 0.44 amps
For a bulb consuming 100 watts:
R = 144 ohms
I = 0.83 ampsFor a 250 watt bulb:
R = 58 ohms
I = 2.1 ampsThe burn-in current should be limited to no more than around 1/3 or at most 1/2 of the current rating of the fuse.
Variacs intended for 120 volt applications can generally be turned up to put out voltages that are somewhat higher than that, so you would probably want to use a multimeter to determine the setting corresponding to 120 volts.
... Regarding flicker bulbs, a point to keep in mind is that the only audio components whose AC power draw fluctuates significantly are power amplifiers that are not biased in class A. Or subwoofers or speakers which incorporate such amplifiers. Class A amplifiers, preamps, other line-level components, phono stages, etc., all draw essentially the same amount of AC current at all times.
Regards,
-- Al