Haryana PMT Haryana PMT Solved Paper-2001

  • question_answer
    A bulb has specifications of 1 kW and 250 volt, the resistance of bulb is :

    A)  625 \[\Omega \]                             

    B)  0.25 \[\Omega \]

    C)  6.25 \[\Omega \]                            

    D)  62.5 \[\Omega \]

    Correct Answer: D

    Solution :

                    From     \[P=\frac{{{V}^{2}}}{R}\] or         \[R=\frac{{{V}^{2}}}{P}\]               ...(1) Here, \[V=250volt,\text{ }P=1kW=1000\text{ }watt\]. From equation (1),                 \[R=\frac{250\times 250}{1000}=62.5\Omega \]


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