Electronic – Risk in limiting current with light bulbs in parallel

The circuit shown will work, but the transistor and Rsense create a voltage drop that must be taken into account.
What you are seeing is the effect of this:

At 480mA the voltage drop across the 10Ω resistor would be 4.8V, which leaves no "room" for the transistor saturation voltage or the Rsense voltage drops.
So the current will be (Vsupply - Qsat - Vrsense) / Rload. To fix this, raise the supply by a couple of volts and try the 0Ω and 10Ω tests again. Also, lower Rdefend considerably (<10Ω)
You should hopefully see (almost) no difference.

For better results, the more gain you have the better. Another thing to note is (as Dave mentions in his answer) that Rbias needs to have a higher limiting point than the Rsense setting, otherwise it will dominate. If the transistor has a gain of 65, and you want Rsense set for 500mA, then Rbias must be set to allow more than 500mA. At 500Ω, it will set the absolute limit at 65 * ((5V - 1.4V) / 500Ω) = 468mA, so even if Rsense were set for 500mA you won't get it. To avoid this set Rbias for e.g. 250Ω, or as mentioned below use a MOSFET for Q1 and then the value isn't as important (10kΩ will do)

Another option is to use a common opamp constant current circuit:

Simulation with a supply of 4.8V, current limited to 500mA, Rload swept from 1mΩ to 50Ω and current through it plotted relative to this (note current stays flat at 500mA whilst limited):

This meets your requirements of a solid 500mA limiting at 4.8V supply, and is easily adjustable by varying the opamp non-inverting via input voltage R2/R3 divider. The formula is V(opamp+) / Rsense = I(Rload) For example, the 1V reference is divided by 20 to provide 50mV at the opamp+ input, so 50mV / 100mΩ = 500mA.
A MOSFET is used to avoid base current errors complicating matters (a MOSFET with low Vth can also be used in the original transistor circuit to improve things)

The advantage of an incandescent light bulb over a resistor is that its cold resistance is about 10 times less than when lit up to normal brightness. Therefore you can choose a wattage that limits short circuit current to a safe value, without excessive voltage drop at lower currents. Also, since it takes some time for the filament to heat up and increase resistance, short duration current surges cause much less voltage drop.

You should choose a lamp with voltage rating close to the power supply voltage (a little lower won't hurt) and wattage that gives an acceptable short circuit current. Power = Volts x Amps, so if you wanted to limit a 12V supply to 0.5A then you would need a 12V 6W lamp.

The maximum normal operating current will then be determined by how much voltage sag is acceptable. This is tricky to calculate because the lamp's change in resistance with current is very non-linear. A 12V 15W festoon lamp that I tested had a resistance of 0.9 Ohms when cold, and 9.8 Ohms at full brightness. At 0.3A (24% of the bulb's nominal current draw) the output voltage dropped from 12.0V to 11.5V. Here is a graph showing output voltage vs current for my lamp compared to using a 10 Ohm resistor.

The 'soft' current limiting that a lamp provides is good for protecting against continuous short circuits without affecting normal operation, but won't protect sensitive electronic components which cannot withstand short current surges or slightly higher than normal operating current.

If you need a fast acting current limit with sharper cutoff then you could use a bipolar transistor (PNP for positive voltage, NPN for negative) with its Emitter connected to the power supply, Collector to the load and a high value resistor going from Base to Ground. Maximum collector current will then be determined by the Base current multiplied by the transistor's current gain (which varies from unit to unit, so you may have to adjust the resistor value to get a precise current limit).