Ohm's law applies and those bulbs are indeed just simple resistive loads. To get the series current, first derive the equivalent resistance at rated voltage and current for each lamp type. Add up these resistances and from that and the applied voltage to the entire segment derive the series current.
The bulbs with a lower voltage rating will most likely shine brighter than the higher rated ones. Since both types are in series, the same current passes through them. However, the energy dissipated (in the form of light and heat) by the filament may be different.
From an application note for aerospace lamp components:
Design Volts: The voltage for which the lamp was originally designed.
All other ratings (amperes, brightness, and life) are measured at this
voltage, and can be changed by rerating the voltage.
...
Candlepower is directly proportional to the 3.5 power of the ratio
of the applied voltage, and can be increased at the expense of lamp
life.
The question aroused my interest enough to set up an experiment. I changed the question's parameters in one key aspect: Instead of an LED strip with multiple LEDs in series, I hooked up 3 blue LEDs (Vf = ~2.8 Volts each) in parallel, with a single 100 Ohm resistor to limit current to all 3, to a 0.047 Farad, 5.5 Volt coin type "motherboard supercap".
I know, sharing a resistor is really bad practice, so just use separate resistors for your own experiment.
The supercap was charged from a pair of AA alkaline cells (~3.12 Volts across capacitor after 3 minutes), then the wires to the battery were pulled out.
simulate this circuit – Schematic created using CircuitLab
While the dimming effect was an expected outcome, the results were startling: The LEDs stayed lit at diminishing intensity for over a minute after disconnecting the battery. Here is the video I took of the experiment.
The reason the LEDs stayed lit so much longer than expected is that a typical LED continues to be illuminated down to well under 5% of its nominal current - In the case of the LEDs I used, at around the 1 minute mark they were quite visible, if dim, with a mere 1 mA split between all three.
The LEDs finally dimmed to nothingness after perhaps 15 minutes.
Conclusions:
- A much smaller capacitance than the 0.047 Farad supercapacitor used here would be preferred for the purpose envisaged.
- If one must use a 12 Volt 20 mA LED strip, instead of LEDs in parallel, then a set of 3 of these coin supercaps in series would work: The resultant capacitance of around 0.0157 Farad will provide a dimming duration closer to the OP's target of 2 to 10 seconds, instead of the unbearably long 1 minute dimming observed in the video.
- The reason some previously posted capacitance calculations including my own 0.5 Farad comment were far off the mark is because the reducing current flow due to discharge, i.e. the very dimming effect being sought, was unaccounted for.
- For any comments that might arise about the "unacceptably high" ESR of these motherboard supercaps, it is clear that theory needs to be backed up by practical experimentation, as done for this answer.
The supercapacitor I used is sold for under $2 a pair, including international shipping, on eBay:
Not quite the tens or hundreds of dollars that I, and others, had previously mentioned.
Addeddum thanks to discussion with @DavidKessener:
- If using multiple supercaps in series and charged to a higher voltage for the string, than the individual capacitor's rated voltage, biasing resistors are required to prolong the life of the capacitors. Without these, the capacitors will charge unevenly, and will eventually die faster.
- Based on this Maxwell appnote, and taking a leakage current per capacitor of 10 uA (the actual leakage current of these particular caps is much lower, so even safer), we get a 55 kOhm value for biasing resistors to pass
10 x 10 = 100 uA
, so add 3 new 56k resistors as below, for using a 12 Volt supply and a 12 Volt LED strip
simulate this circuit
Best Answer
You could easily connect them all in parallel. or in any grid ie (3*10 or 6*5) and just control the input to the grid independently. As others have said incandescent bulbs is mostly a resistive load. If you go with ,say 5 in series and each of those 5 is in parallel then you would require 7.5V @ 1.8A. Choosing a configuration is simply a matter of what you have available.
Except for requiring more power than LEDs the incandescent bulbs can be pwm much easier. You might even be able to get away with a much lower switching frequency because the bulbs will tend to filter higher frequencies.
Series is easier to switch m but if one bulb do fails , the whole light fails. Going parallel you need to switch a higher current but its trivial to find a blown bulb. Using a grid like arrangement mixes these two features partially so you get lower switching current than when all is in parallel and its easier to find faults than when everything is in series.