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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