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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    If a 30V, 90W bulb is to be worked in 120V line, the resistance to be connected in series with the bulb is:

    A)  \[20~\,\Omega \]                          

    B)  \[10~\,\Omega \]

    C)  \[40~\,\Omega \]                          

    D)  \[30~\,\Omega \]

    Correct Answer: D

    Solution :

    Here, resistance of bulb                 \[{{R}_{1}}=\frac{{{V}^{2}}}{P}\]                 \[=\frac{30\times 30}{90}=10\,\,\Omega \] current in the bulb \[i=\frac{P}{V}=\frac{90}{30}\]                                 = 3 amp Let resistance \[x\] is put in series when bulb is connected to \[V=120\] volt supply. Then                 \[3=\frac{120}{10+x}\] or            \[x=30\,\,\Omega \]


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