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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    When two resistances \[{{R}_{1}}\] and \[{{R}_{2}}\] are connected in series, they consume 12 W power. When they are connected in parallel, they consume 50 W power. What the ratio of the powers of\[{{R}_{1}}\]and\[{{R}_{2}}\]?

    A)  1/4                       

    B)  4

    C)  3/2                       

    D)         3

    E)  1

    Correct Answer: C

    Solution :

    In series, \[{{p}_{s}}=\frac{{{P}_{1}}\times {{P}_{2}}}{{{P}_{1}}+{{P}_{2}}}=12W\] In parallel,   \[{{P}_{p}}={{P}_{1}}+{{P}_{2}}=50\,W\] \[\therefore \]  \[{{P}_{1}}{{P}_{2}}=12\times 50=600\] Now,     \[{{({{p}_{1}}-{{p}_{2}})}^{2}}={{({{p}_{1}}+{{p}_{2}})}^{2}}-4{{p}_{1}}{{p}_{2}}\] \[={{(50)}^{2}}-4\times 600\] \[=2500-2400=100\] \[\therefore \]  \[{{P}_{1}}-{{P}_{2}}=10\] or            \[{{P}_{1}}=30W,{{P}_{2}}=20\,W\] \[\therefore \]  \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{3}{2}\]


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