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  3. CET Karnataka Medical

CET Karnataka Medical CET - Karnataka Medical 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 \[i=\frac{P}{V}=\frac{90}{30}=3amp\] Let resistance x is put in series when bulb is connected to Y = 120 volt supply. Then                \[3=\frac{120}{10+x}\] or            \[x=30\Omega \]


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