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BHU PMT BHU PMT (Screening) Solved Paper-2006

  • question_answer
    In a series LCR circuit the frequency of a 10 V, AC voltage source is adjusted in such a fashion that the reactance of the inductor measures \[15\,\Omega \]and that of the capacitor\[11\,\Omega \]. If\[R=3\,\Omega \]the potential difference across the series combination of Z, and C will be:

    A)  8V                                         

    B)  10 V

    C)  22V                                      

    D)  52V

    Correct Answer: A

    Solution :

                     Key Idea: The potential difference across the series combination of L and C is the difference in potential differences across individuals. Current through the circuit,                 \[i=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{c}})}^{2}}}}\] Given,\[R=3\,\Omega ,\,{{X}_{L}}=15\,\Omega ,\,{{X}_{C}}=11\,\Omega ,\] \[{{V}_{rms}}=10\,V\] \[\therefore \]  \[i=\frac{10}{\sqrt{{{(3)}^{2}}+{{(15-11)}^{2}}}}\] Or           \[i=\frac{10}{\sqrt{9+16}}\] Or           \[i=\frac{10}{5}=2A\] Since L, C and R are connected in series combination, then potential difference across R is                 \[{{V}_{R}}=iR=2\times 3=6V\] Across \[L,{{V}_{L}}=i{{X}_{L}}=2\times 15=30V\] Across \[C,{{V}_{C}}=i{{X}_{C}}=2\times 11=22V\] So,   potential   difference across series combination of L and C                 \[={{V}_{L}}-{{V}_{C}}\]                 \[=30-22\]                 \[=8\,V\] Note: In\[LCR\]circuit whenever voltage across various elements is asked, find rms values unless stated in the question for the peak or instantaneous value.


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