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BCECE Medical BCECE Medical Solved Papers-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 Q and that of capacitor 11\[\Omega \]. If R = 3\[\Omega \], the potential difference across the series combination of L and C will be :

    A)  8V            

    B)  10 V

    C)  22V           

    D)  52V

    Correct Answer: A

    Solution :

    Given, \[R=3\,\Omega ,\,{{X}_{L}}=15\,\Omega ,\,{{X}_{C}}=11\,\Omega \],                                  \[{{V}_{R}}=10\,V\] \[\therefore \] Current through the circuit                 \[i=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}\]                 \[=\frac{{{V}_{rms}}}{\sqrt{{{(3)}^{2}}+{{(15-11)}^{2}}}}\]                 \[=\frac{10}{\sqrt{9+16}}\]                 \[=\frac{10}{5}=2\,A\] Since, L, C and R are connected in series combination the potential difference across R is                 \[{{V}_{R}}=i\times R=2\times 3\]                 = 6 V Across L,                 \[{{V}_{L}}=i{{X}_{L}}=2\times 15=30\,N\] Across C,              \[{{V}_{C}}=i\,{{X}_{C}}\]                                 \[=2\times 11=22\,V\] So, potential difference across series combination of L and C is                 \[={{V}_{L}}-{{V}_{C}}\] \[=30-22=8\,V\]


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