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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    If\[{{R}_{1}}\]and\[{{R}_{2}}\]be the resistances of the filaments of 200 W and 100 W electric bulbs operating at 220 V, then\[\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)\]is

    A)  1                            

    B)         2                            

    C)  0.5                        

    D)         4

    E)  0.25

    Correct Answer: C

    Solution :

    Power of bulb, \[P=\frac{{{V}^{2}}}{R}\] \[\therefore \]  \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\] Or           \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{100}{200}=0.5\]


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