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BVP Medical BVP Medical Solved Paper-2002

  • question_answer
    An electric bulb of 100 W is connected to a supply of electricity of 200V. Resistance of the filament is :

    A)  242 \[\Omega \]                             

    B)  22000 \[\Omega \]

    C)  100\[\Omega \]                              

    D)  484 \[\Omega \]

    Correct Answer: D

    Solution :

                    \[P=100\text{ }W,\text{ }V=220\text{ }volt\] \[P=\frac{{{V}^{2}}}{R}\]     \[\Rightarrow \]  \[R=\frac{{{V}^{2}}}{P}\] \[=\frac{220\times 220}{100}=484\Omega \]


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