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BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    What is the resistance of a 40 W lamp which is lighted at full brilliance by a current of \[\frac{\text{1}}{\text{3}}\text{A}\]?

    A)  \[450\,\Omega \]                           

    B)  \[360\,\Omega \]                           

    C)  \[120\,\Omega \]           

    D)         \[13.33\,\Omega \]

    Correct Answer: B

    Solution :

    Power of electric bulb is given by \[P=\frac{{{V}^{2}}}{R}\] But from Ohms law, \[V=iR\] So,          \[P=\frac{{{(iR)}^{2}}}{R}={{i}^{2}}R\] \[\therefore \]  \[R=\frac{P}{{{i}^{2}}}\] Here,                     \[P=40\,W,\,i=\frac{1}{3}A\] Hence, \[R=\frac{40}{{{(1/3)}^{2}}}=40\times 9=360\,\Omega \]


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