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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    An alternating voltage\[e=200\text{ }sin\text{ }100t\]is applied to a series combination\[R=30\,\Omega \]and an inductor of 400 mH. The power factor of the circuit is

    A)  0.01                      

    B)         0.2                        

    C)  0.05                      

    D)         0.042

    E)  0.6

    Correct Answer: E

    Solution :

    Power factor \[\cos \phi =\frac{R}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\]                 \[=\frac{30}{\sqrt{{{(30)}^{2}}+{{(100)}^{2}}\times {{(400\times {{10}^{-3}})}^{2}}}}\]                 \[=\frac{30}{\sqrt{900+1600}}=\frac{30}{50}=0.6\]


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