CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    An electric bulb rated 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 W is

    A)  \[10\,\Omega ,\]           

    B)         \[15\,\Omega ,\]

    C)  \[2.5\,\Omega ,\]           

    D)         \[25\,\Omega ,\]

    E)  \[20\,\Omega ,\]

    Correct Answer: E

    Solution :

    \[P=\frac{{{V}^{2}}}{R}\] \[500=\frac{10000}{R}\] \[R=20\,\Omega \] \[P={{i}^{2}}R\] \[{{i}^{2}}=\frac{500}{R}=\frac{500}{20}=25\] \[i=5A\] \[\therefore \]  \[i=\frac{V}{(R+{{R}_{1}})}=\frac{200}{(20+{{R}_{1}})}\]                 \[5=\frac{200}{(20+{{R}_{1}})}\] \[20+{{R}_{1}}=40\]                 \[{{R}_{1}}=20\,\Omega \]


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