It's been a week since a light bulb in the bathroom died. I'm wondering if the bulb is dead, does it still consume electricity if the switch is in ON?
The light is completely dead, no light whatsoever.
It depends on the type of bulb.
Regular incandescents won't consume any electricity if the bulb is dead, since there's no continuous path for the current to take. It's just like an open switch.
With CFLs and LEDs, it depends on why the bulb burned out, but in general they will consume some amount of electricity even when burned out. Some CFLs may even consume up to 50% as much as a good bulb (older link, but a lot of burned out bulbs may be old). Newer bulbs may have circuits which eliminate most electricity usage on dead bulbs, as this answer from the electronics stack shows.
Smart bulbs have additional electronics, and so would consume even more electricity than an equivalent non-smart bulb, assuming of course that it's not the smart electronics that died.
The only way to be sure is to measure the usage, with a device like a Kill-a-Watt meter. You would need to install the bulb in a lamp or other fixture with a plug.
If it is a true incandescent light bulb: NO, apart from very very minor losses (through insulation imperfections and transmission line effects) due to the fact that a longer run of wiring is now live.
In addition, if it is an old school flourescent fixture: Very minor losses due to EMI filtering circuitry.
In addition, if it is a LED bulb using a capacitor-based passive power supply: Depends on how the LEDs itself failed. LEDs can fail in a way that they still generate heat (or even pose a short circuit, which would put all the energy into the current limiting circuitry) but no light.
In addition, if there is any active electronics inside (modern LEDs or CFLs), it depends on how these failed and/or react to failure of the actual lighting component - no general statement possible without knowing the exact circuitry.
Nope. Basic principles of electricity: electrical current doesn't flow through an open circuit (at least not at the voltages a residence sees). When a bulb burns out, the conductive path through the bulb is broken and the circuit becomes open - effectively an infinite load. Same as if a breaker were to open.
Well we've learned one thing the word depends is the most commonly used verb in our industry.
Incandescent of course not.
Any lamp using a magnetic ballasts (oldstyle for any fluorescent or low and high pressure gas lamps) is an autotransformer and power will pass through it even though there is no load and like mmathis has said could be up to 50%.
Newer Electronic ballasts and drivers for LED's have the ability to sense whether or not there is a load and shut down. So if all lamps burn out, it will use some trace power but not enough where I would be concerned about usage.
There is one scenario I don't see mentioned: Christmas tree lights.
Between the eras of the old large bulb Christmas lights and the "modern" LED lights there was a period of small incandescent lights. Typically 10-30 small incandescent bulbs would be strung in series on a string (sometimes with several series strings physically assembled into a longer string).
Since the lifetime of these tiny bulbs was unpredictable, and since if any bulb in a series goes out the entire string goes out, a technique was developed to tolerate a few dead bulbs in a string.
Basically, at in the base of each bulb was a small glob of conductive material with carefully chosen characteristics. If you had a 10-bulb string of 12-volt bulbs, for 120v total, the conductive glob would only draw a small amount of current and not get very hot. But if the filament of a bulb burned out then almost the entire 120v would be applied across the conductive glob and it would carry 10 times the current and (if everything went as to plan) get hot enough to "melt" (change phases somehow). When it melted, it's resistance would drop to near (but not quite) zero, and the defective bulb would be effectively shorted out.
So a Christmas tree lamp of this style can be consuming a small amount of power when "burned out", whether the "glob" is "melted" or not.