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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    When a DC voltage of 200 V is applied to a coil of self-inductance\[\left( \frac{2\sqrt{3}}{\pi } \right)H,\] a current of 1 A flows through it. But by replacing DC source with AC source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of AC supply is

    A)  100 Hz                                 

    B)  75 Hz   

    C)         60 Hz                   

    D)         30 Hz

    E)  50 Hz

    Correct Answer: E

    Solution :

    Resistance of coil\[(R)=\frac{200}{1}200\,\Omega \] Current         \[I=\frac{200}{\sqrt{{{R}^{2}}+X_{L}^{2}}}\] Or           \[0.5=\frac{200}{\sqrt{{{R}^{2}}+X_{L}^{2}}}\] Or           \[{{R}^{2}}+{{(2\pi fL)}^{2}}={{(400)}^{2}}\] Or           \[or{{\left( 2\pi f\times \frac{2\sqrt{3}}{\pi } \right)}^{2}}={{(400)}^{2}}-{{(200)}^{2}}\]                                                 \[=120000\] Or           \[4f\sqrt{3}=200\sqrt{3}\] Or           \[f=50\,Hz\]      


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