If 2 household light bulbs rated 60 and 100W are connected in series to household power...?

then which will be brighter? And why?

Answer:
If they are in series, the current through each must be equal, and since the resistance of the 60 watt bulb is higher, more power will be dissipated in it, thus making it brighter

EDIT:
Since there are conflicting opinions of which bulb has higher resistance. Consider each bulb on it's own. Assuming 120V, the 60V bulb draws .5A, 120 x .5 = 60, for .5A, the resistance must be 120/.5 = 240 ohm. The 100W, 100/120 = .833A 120/.833 = 144 ohm
or, in one step, 120²/60 = 240, 120²/100 = 144

The 100 watt will burn brighter due to the fact it provides more resistance.

E=iR, or i=E/R
P=iE and substituting i from above get P=E^2/R

Power equals voltage squared divided by resistance.

So, now calculate the R for th 60W bulb:
60 = (120*120)/R or R= (120*120)/60 = 240 ohms

And for the100W:
100= (120*120)/R or R=(120*120)/100 = 144 ohms

The total in series is 240ohms +144ohms = 384 ohms

The current through 384 ohms at 120 volts is:

i=E/R or i = 120/384 or .3125 amps

Now, substitute E=iR into P=iE and get P=i^2R
Power is current squared times resistance.

for the 60 watt bulb: P=.3125 *.3125* 240 = 23.4375 Watts

for the 100 watt bulb: P= .3125*3125*144 =14.0625 Watts

The power is proportional to brightness---

So the 60 watt bulb will be brighter!

Ok well I'm not a freakin engineer but I'm the greatest genius that ever lived so I feel competent to include my opinion. If it's dc the 60 watt for reasons stated. If it's ac the 100 watt since ac power provides energy to all energy draws equally.

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If 2 household light bulbs rated 60 and 100W are connected in series to household power...?

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