KVPY Sample Paper KVPY Stream-SX Model Paper-21

  • question_answer
    In an LRC series circuit at resonance, current in the circuit is \[10\sqrt{2}A\]. If now frequency of the source is changed such that now current lags by \[45{}^\circ \] than applied voltage in the circuit. Which of the following is correct is -

    A) Frequency must be increased and current after the change is 10 A

    B) Frequency must be decreased and current after the change is 10 A

    C) Frequency must be decreased and current is same as that of initial value

    D) The given information is insufficient to conclude anything

    Correct Answer: A

    Solution :

    Initially at resonance: \[{{X}_{L}}={{X}_{E}}\Rightarrow Z=R.\]
    \[\therefore {{i}_{0}}=\frac{{{\varepsilon }_{0}}}{R}=10\sqrt{2}\,A\]
    After increasing frequency: \[{{X}_{L}}>{{X}_{c}}\]\[\omega L>I/\omega C\]
    \[\omega >\frac{1}{\sqrt{LC}}\Rightarrow \omega >{{\omega }_{0}}\] (i.e. frequency increases) and \[i=\frac{{{\varepsilon }_{0}}}{\sqrt{{{R}^{2}}+{{X}^{2}}}}=\frac{{{\varepsilon }_{0}}}{\sqrt{2R}}={{i}_{0}}/\sqrt{2}\]\[=10\,\,amp.\]


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