College Physics
posted by Kate .
The power dissipated by a light bulb increases by 50 W when the potential difference applied across the light bulb increases from 50 V to 100 V. Find the resistance of the light bulb.

p=50^2/R
P+50=(100)^2/R
subtract the first equation from the second
50=1/R (100,0002500)
solve for R. Check my thinking. 
It worked! :D