Homework Help: College Physics

Posted by Kate on Sunday, February 7, 2010 at 12:15am.

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.

College Physics - bobpursley, Sunday, February 7, 2010 at 7:57am
p=50^2/R
P+50=(100)^2/R
subtract the first equation from the second
50=1/R (100,000-2500)
solve for R. Check my thinking.
College Physics - Kate, Wednesday, February 10, 2010 at 4:12pm
It worked! :D

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