Asked 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.
Answers
Answered by
bobpursley
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.
P+50=(100)^2/R
subtract the first equation from the second
50=1/R (100,000-2500)
solve for R. Check my thinking.
Answered by
Kate
It worked! :D
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