In my physics class I have this problem that shows two lightbulbs, one 60W and one 100W in series, connected to a 120V battery. The problems are:
Which bulb is brighter? (A: 60W)
Calculate the power dissipated by the 60W bulb. (A: 23.4W)
Calculate the power dissipated by the 100W bulb. (A: 14.1W)
Why is the power dissipated not simply the wattages of the bulbs? I followed one workthrough online where you first find R for both using P = (V^2)/R and then use I = V/R to get a current of 0.3125A. The power dissipated is then calculated using P = I^2R and you get the above answers. However, doesn't that assume the voltage drop between the lightbulbs is 120V in both cases, and isn't that wrong?
I tried getting it another way where I said P1 = IV1, P2 = IV2, and V1+V2=120(Volts). I solved the voltage drop on the 60W lightbulb to be 45V and 75V on the 100W one. Then, current is solved to be 4/3A which lets us solve the resistance for each one as 33.75Ohms and 56.25Ohms. Then using the formula P = V^2/R, the original wattages are found as the answer. Why is it right to assume 120V for both bulbs?