Only the master electrician would know

[aoyuki]

a master electrician would never wire these in series

[timmathieu]

Suppose this is a 120V circuit. The 40W bulb would have a resistance of 360 ohms (ohms law: volts squared divided by power equals resistance) Each 5W bulb would have a resistance of 2, 880 ohms. Since they are all in series, you have a total circuit resistance of 6, 120 ohms. 120/6120=.02 amps.

Now you just multiply amps times resistance of each bulb to find the voltage drop across each one. Both 5W bulbs will have 57.6V across them and the 40W bulb will have 7.2V across them.

Chances are, the CFLs will not operate properly at 57.6V across them. The middle incandescent MIGHT show a little glow at 7.2V.

[egoruderico]

Did you try turning it OFF and then ON again?

[AzeveidoMateus]

This is one reason why residential lighting is wired in parallel not series.

[marcoromanelli]

A master eletrician should know not to wire a circuit like that...

[briansmith]

I'm a union electrician so I'll give you an answer after my break is over

[chrishettinger]

Different types of bulbs mean different driver circuits. This cannot be resolved into equivalent resistance. The answer is it is dependent upon the electronics "under the hood".

[johnjohn-nefw]

After a considerable amount of time thinking this over i have determined that the bulb must be replaced

[desislavarashev]

In a series circuit, all bulbs must have the same current flow, but the 40W bulb requires more current than the 5W bulbs can handle, so it doesn't turn on.

[DeadKoby]

An incandescent bulb in series /w/ your power line is an electronics nerd trick to limit current. The incandescent bulb is passing current, but the other bulbs aren't drawing enough current to make it light.

[DilpreetSingh-swei]

In series voltage is divided among each element, but current remains the same.
If they were in parallel, it might work.

[muhammadsawleh]

Voltage drops in series as the load is of 40W so their is not enough voltage to power up the bulb.

[drmantistoboggan]

Dunno mate, whole system must be buggered. I recommend replacing all of it from the ground up.

[Shivani]

Resistance of the 40 watt bulb is more than 5 watt so when the resistance is more than the current is less and the voltage is less

[BennyHueENT]

Correct if wrong but I'm pretty sure the first 5 is like a resister and lowers output current to what the bulb can handle so 5w worth of current passed through the 40 can't heat the filament enough to glow it's "powered" but not lit and goes right on to the 5 with enough to light it and back to source

[martinvkumar]

Thank goodness, everyone here is electrical engineers.

[MrDaniyalAh]

Let's assume numbers and figure this out.
220V
so 5W and 40W bulbs resistance is 9690 ohms and 1210 ohms respectively (R=V^2/P)
All three in series is 20590 ohms. At 220V this means 9.74mA will flow. (I=V/R)
So calculating the power that each bulb draws in this situation: for 5W it's 0.917W and for 40W it's 0.115W (P=I^2 R)
Since the 5W bulbs are getting 18% of thier rated power they will light up dimly, the 40W one is getting only 0.3% of its rated power so it doesn't light up at all.
Am an electrical engineer, I do things like this mentally multiple times a day.

[PlaywithJunk]

The 5W lamps act like large resistors. They do not allow enough current to flow for the filament bulb to get hot enough. Therefore no light...

[mts]

Because the current is passing in the circuit.The 40w bulb does not light due to the poor or high resistance on those CFL lamps

[cornelisemail]

The 2 5 watts powersaver bulbs are working as a resistance...
The 40 watt is now just a dead normal wire.