Power dissipated by resistor -- Right answer, but need insight

[kostoglotov]

Right, but the value of that current is not calculated in the way you calculated it. You assumed that the bulb with the higher resistance is all you need consider when calculating that current. You must instead consider both bulbs.

Ok, so I'd add their currents to get their total currents? So $I_{total} = V(\frac{1}{R_1}+\frac{1}{R_2})$, and then use $I_{total}$ to calculate the power use of each "resistor"?

 

[Herman Trivilino]

Ok, so I'd add their currents to get their total currents? So $I_{total} = V(\frac{1}{R_1}+\frac{1}{R_2})$, and then use $I_{total}$ to calculate the power use of each "resistor"?

No, you add their voltage drops to get the total voltage drop. (The currents are the same, if you added them together you'd get 2$I$.).

$I = \frac{V}{R_1+R_2}$

 

[haruspex]

I'll never forget embarrassing myself by assuming the bulbs have a resistance that stays the same when you hook them up in series. They don't, by a long shot.

True, but at the level of the present discussion...