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An electric bulb rated as 200 \mathrm{~W} at 100 \mathrm{~V} is used in a circuit having 200 \mathrm{~V} supply. The resistance ' R ' that must be put in series with the bulb so that the bulb delivers the same power is ________\Omega.
 

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best_answer

     


200 W,100 V
R_{Bulb}= \frac{V^{2}Rating}{P_{Rating}}= \frac{104}{200}
R_{Bulb}=50
For power across the bulb to be 200 W, the voltage drop must be 100 V
So, the remaining 100 v is across R
Since the current is the same through resistance R and Bulb
V= V_{Bulb}= 100
IR= IR_{Bulb}
R= R_{Bulb}= 50\Omega

 

 

Posted by

vishal kumar

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