Question

Light bulb packages will have at least two numbers on them; a power and what the potential difference across the bulb must be so that the bulb has the listed power. You go to the store and purchase two light bulbs. The package for Bulb A says 40 W in a large font with 120 V in a smaller font; the package for Bulb B says 60 W in a large font with 120 V in a smaller font. You then connect the light bulbs as shown below (where the emf is 120 V):
http://i.imgur.com/R37jUsf.png
Which bulb is brighter and what is its power dissipation?

Answer 1

Assume that the resistance of the bulbs is independent of the current.
Resistance of 40 W bulb =\[\frac{V ^{2}}{W}=\frac{120^{2}}{40}=360\ ohms\]
Resistance of 60 W bulb =
\[\frac{120^{2}}{60}=240\ ohms\]
Total resistance of bulbs when connected in series = 360 + 240 = 600 ohms.
Current through both lamps in series = 120/600 = 0.2 A
\[Power=I ^{2}\times R\]
Therefore in the series connection the lamp with the larger resistance will dissipate the higher power. In this case the lamp marked 40 W will be brighter and its power dissipation will be
\[0.2^{2}\times 360=14.4\ W\]