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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    An ideal choke draws a current of 8 A when connected to an AC supply of 100 V, 50 Hz. A pure resistor draws a current of 10 A when connected to the same source. The ideal choke and the resistor are connected in series and then connected to the AC source of 150 V, 40 Hz. The current in the circuit becomes

    A)   \[\frac{15}{\sqrt{2}}A\]                                

    B)   8 A

    C)   18 A                                      

    D)   10 A

    Correct Answer: A

    Solution :

     Resistance, \[R=\frac{100}{10}=10\,\Omega \] Inductive reactance, \[{{X}_{L}}=2\pi fL\]                 \[\frac{100}{8}=2\pi \times 50\times L\] \[\Rightarrow \]               \[L=\frac{1}{8\pi }H\]                 \[X{{'}_{L}}=2\pi f'L=2\pi \times 40\times \frac{1}{8\pi }=10\,\Omega \] Impedance of the circuit is \[Z=\sqrt{{{R}^{2}}+X_{L}^{'2}}\]                   \[=\sqrt{{{(10)}^{2}}+{{(10)}^{2}}}=10\sqrt{2}\,\Omega \] Current in the circuit is \[i=\frac{V}{Z}=\frac{150}{10\sqrt{2}}=\frac{15}{\sqrt{2}}A\]


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